JavaScript (Node.js), 50 bytes

f=(n,p=1e3)=>n<p?(n+p+'').slice(1):'#'+f(n-p,p/10)

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TIO was based on Arnauld's answer. Show 👑 as #.

JavaScript (ES6), 62 bytes

Outputs crowns as x characters.

n=>(n+1e3+'').replace(/^21*0/,s=>s.replace(/./g,`x`)).slice(1)

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How?

We add $1000$ to the input, coerce it to a string, look for the /^21*0/ pattern and replace each matching character with x. Finally, we remove the leading character.

Examples:

$$beginalign&0 &rightarrow &text "1000" &rightarrow &text "1000" &rightarrow &text "000"\
&123 &rightarrow &text "1123" &rightarrow &text "1123" &rightarrow &text "123"\
&1023 &rightarrow &text "colorredtext20text23" &rightarrow &text "xx23" &rightarrow &text "x23"\
&1103 &rightarrow &text "colorredtext210text3" &rightarrow &text "xxx3" &rightarrow &text "xx3"\
&1110 &rightarrow &text "colorredtext2110text" &rightarrow &text "xxxx" &rightarrow &text "xxx"
endalign
$$

Shakespeare Programming Language, 763 692 690 689 bytes

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy!Ajax:You big big cat.Scene V:.Ajax:Remember the remainder of the quotient between I twice the sum of a cat a big big cat.Ford:You be the quotient between you twice the sum of a cat a big big cat.Ajax:You be the sum of you a pig.Be you nicer zero?If solet usScene V.Ford:You big big cat.[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum of you a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

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Uses " " instead of crowns. At the cost of 4 more bytes, this could be modified to show a "visible" character instead.

Explanation:

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]

 Boilerplate, introducing the characters.

Ford:Listen tothy!

 Input a value to Ajax.

Ajax:You big big cat.

 Set Ford's value to 4 (we will be pushing 4 digits from Ajax onto Ford's personal stack).

Scene V:.Ajax:Remember the remainder of the quotient between I twice the sum of a cat a big big cat.Ford:You be the quotient between you twice the sum of a cat a big big cat.

 DIGIT-PUSHING LOOP: Push Ajax's last digit onto Ford's stack; divide Ajax by 10.

Ajax:You be the sum of you a pig.Be you nicer zero?If solet usScene V.

 Decrement Ford; loop until Ford is 0.

Ford:You big big cat.

 Set Ajax's value to 4 (we will pop 3 digits from Ford's stack in the next loop).

[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.

 Pop the top value off Ford's stack, and store that into Page.
 Here, Page will contain 0 if there are no crowns to be drawn,
 and 1 if there are crowns to be drawn.

Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.

 DIGIT-DRAWING LOOP: Pop the top value off of Ford's stack and set Ford equal to that value.
 If there are no crowns to be drawn, output Ford's literal value here, and skip the crown-drawing section.

Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.

 Draw crown.
 If we are drawing crowns, and Ford contains 0 here, then we are now done drawing crowns, and thus we store 0 into Page.
 (Put in one more "big" for the crown to look like an @ symbol.)

Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum of you a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

 Decrement Ajax; loop until Ajax is 1 (i.e. 3 times).

Python 2, 75 bytes

Shamelessly steals Arnauld's regex. Uses C as crown.

lambda k:re.sub("^21*0",lambda g:len(g.group())*"C",`k+1000`)[1:]
import re

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Python 2, 51 bytes

This instead implements tsh's recursive method. Saved 2 bytes thanks to ovs.

f=lambda n,p=1000:n/p and'C'+f(n-p,p/10)or`n+p`[1:]

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Retina 0.8.2, 41 bytes

b((.)|..)b
$#2$*00$&
T`d`_#`(?=....)1+0

Try it online! Uses #s instead of 👑s. Link includes test cases. Explanation:

b((.)|..)b
$#2$*00$&

Pad 1- and 2-digit numbers to three digits.

T`d`_#`(?=....)1+0

Change leading 1s of 4-digit numbers to #s and delete the next 0.

Clean, 87 bytes

Doesn't output crowns (uses c).

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110
=("ccc"+lpad(""<+n rem(10^i))i'0')%(i,9)

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$ n // function $ of `n`
 # i = // define `i` as (the number of digits that aren't crowns)
 3 - // three minus
 n / 1000 - // 1 if first digit is crown
 n / 1100 - // 1 if second digit is crown
 n / 1110 // 1 if third digit is crown
 = ( // the string formed by
 "ccc" + // prefixing three crowns to
 lpad ( // the padding of
 "" <+ n rem (10^i) // non-crown digits of `n`
 ) i '0' // with zeroes
 ) % (i, 9) // and removing the extra crowns

Clean, 99 - 3 = 96 bytes

This one has crowns.

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110
=("👑👑👑"+lpad(""<+n rem(10^i))i'0')%(i*4,99)

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Jelly, 19 bytes - 0 = 19

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ

A full program printing the result using a space character as the crown.

(As a monadic Link a mixed list of integer digits and space characters is yielded)

Try it online! Or see the test-suite.

...maybe a recursive implementation will be shorter.

How?

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ - Main Link: integer, N e.g. 1010 or 10
 È· - literal 1000 1000 1000
< - less than? 0 1
 ¬ - logical not 1 0
 D - to decimal list [1,0,1,0] [1,0]
 ȧ - logical and [1,0,1,0] 0
 0 - literal zero 0 0
 i - first index - call this I 2 1 (0 treated as [0] by i)
 ɓ - new dyadic chain with swapped arguments - i.e. f(N, I)
 ⁶ - literal space character ' ' ' '
 ⁹ - chain's right argument 2 1
 ẋ - repeat [' ',' '] [' ']
 Ḋ - dequeue [' '] 
 ʋ - last four links as a dyad - i.e. f(N, I)
 +È· - add 1000 2010 1010
 D - to decimal list [2,0,1,0] [1,0,1,0]
 Ḋ - dequeue [0,1,0] [0,1,0]
 ṫ - tail from index (I) [1,0] [0,1,0]
 ; - concatenate [' ',1,0] [0,1,0]
 - implicit print " 10" "010"

Python 2, 52 bytes

lambda n:['%03d'%n,'%3s'%`n`.lstrip('1')[1:]][n>999]

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C, 84 bytes

f(n,k)k=3-n/1000-n/1100-n/1110;printf(k?"%s%0*d":"%s","CCC"+k,k,n%(int)pow(10,k));

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Japt, 20 bytes

A naïve (and slightly drunk!) port of Arnauld's solution. Uses " for crown.

U+A³ s r"^21*0"_çQÃÅ

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Java 10, 84 83 bytes

n->for(int p=100,t;p>0;n-=t%12*p,p/=10)System.out.printf("%c",t=n/p>9?46:48+n/p);

Port of @tsh' C comment.

Uses . instead of crowns.

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Alternative approach (84 (87-3) bytes):

n->f(n,1000)String f(int n,int p)return n<p?(n+p+"").substring(1):"👑"+f(n-p,p/10);

Port of @tsh' JavaScript's answer.

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05AB1E, 20 18 bytes

₄‹i₄+¦ëTð.;„1 „ :

Uses spaces for crowns.

Try it online or verify all test cases.

Explanation:

₄‹i # If the (implicit) input is smaller than 1000:
 ₄+ # Add 1000 to the (implicit) input
 ¦ # And remove the leading 1 (so the leading zeros are added)
 # i.e. 17 → 1017 → "017"
 ë # Else:
 Tð.; # Replace the first "10" with a space " "
 # i.e. 1010 → " 10"
 # i.e. 1101 → "1 1"
 # i.e. 1110 → "11 "
 „1 „ : # Replace every "1 " with " " (until it no longer changes)
 # i.e. " 10" → " 10"
 # i.e. "1 1" → " 1"
 # i.e. "11 " → " "

APL (Dyalog Unicode), 32 bytes

1e3∘⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10

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Prefix direct function.

Port of @tsh's JS Answer.

How:

1e3∘⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10 ⍝ Main function, arguments ⍵ and ⍺ (⍵ → input, ⍺ → 1000).
 ⍵<⍺: ⍝ If ⍵<⍺
 1↓ ⍝ Drop (↓) the first element (1) of
 ⍕ ⍝ Format (⍕); 'stringify'
 ⍵+⍺ ⍝ ⍵+⍺
 ⋄ ⍝ else
 'C', ⍝ Concatenate (,) the literal 'C' with
 ∇⍨ ⍝ Recursive call (∇) with swapped arguments (⍨)
 (⍵-⍺) ⍺÷10 ⍝ New arguments; ⍵ → ⍵-⍺; ⍺ → ⍺÷10

PHP, 71 bytes

for($n=$argn,$x=1e4;1<$x/=10;$n%=$n<$x?$x/10:$x)echo$n<$x?$n/$x*10|0:C;

prints C for the crown. Run as pipe with -nR or try it online.

Haskell, 48 bytes

('2'#).show.(+1000)
n#(a:x)|n==a='c':'1'#x|1>0=x

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