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I've posted the solution for this problem and I'm trying to understand this.

In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac1n^2$ and $b_n$ to be $frac1900^2$ for the next $900^2$ terms? And then hold $b_n$ to be $frac1n^2$ and $a_n$ to be $frac1810000 ^2$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)

And why do we have to add one to the sum of partial sums?

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum mina_n,b_n$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $mina_n,b_n = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.

To find how many terms we add again, think about the pattern
beginalign*
(1+1) - 1 &= 1^2 \
(5 + 1) - 2 &= 2^2 \
(30 + 1) - 6 &= 5^2 \
(930 + 1) - 31 &= 30^2 \
(865830 + 1) - 931 &= 930^2 \
(x + 1) - 865831 &= 865830^2 \
dotsb
endalign*

Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac1n^2$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series
beginmatrix
a_n = bigl(1 & frac12^2 & frac13^2 & frac14^2 & frac15^2 & frac16^2 & frac17^2 & frac18^2 & frac19^2 & frac110^2 & frac111^2 & frac112^2 & frac113^2 & frac114^2 & frac115^2 & frac116^2 & cdots &bigr) \
b_n = bigl(1 & frac12^2 & frac13^2 & frac14^2 & frac15^2 & frac16^2 & frac17^2 & frac18^2 & frac19^2 & frac110^2 & frac111^2 & frac112^2 & frac113^2 & frac114^2 & frac115^2 & frac116^2 & cdots &bigr)
endmatrix
and modify them like so:
beginmatrix
a_n = bigl(1 & colorredfrac14 & colorredfrac14 & colorredfrac14 & colorredfrac14 & frac16^2 & frac17^2 & cdots & frac129^2 & colorredfrac1900 & colorredfrac1900 & colorredfrac1900 & cdots & colorredfrac1900 & colorredfrac1900 & frac1930^2 & cdots &bigr) \
b_n = bigl(1 & frac12^2 & frac13^2 & frac14^2 & colorredfrac125 & colorredfrac125 & colorredfrac125 & cdots & colorredfrac125 & colorredfrac125 & frac131^2 & frac132^2 & cdots & frac1928^2 & colorredfrac1929^2 & colorredfrac1929^2 & cdots &bigr).
endmatrix
The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac1n^2$.