Anything $0$ or $1\bmod{9}$ works. To achieve the sum of the digits $0\bmod{9}$, consider $n=99\dots9$, and for $1\bmod{9}$, consider $n=199\dots9$. We can easily confirm that no other values work by considering the quadratic residues modulo 9.

First, let's analise things mod 9: $s(n) = s(n^2) \implies n^2\equiv n (9) \implies n \equiv s(n) \equiv \{ 0,1 \} (9)$. We claim that for all $k$, there's a $n$ for which $s(n)=s(n^2)=9k$ or $s(n)=s(n^2)=9k+1$. I've constructed $n=999...9$ for the first case, and I'll prove it works. For the second case, the proof that $1999...9$ is analogous. Notice that if $s(999...9999)=9k$, then $s((999...9999)^2)=9k$ too. Then, we analyse the product: $9999....9$x$9999...9$=$\sum_{k=0}^{n-1}\overline{89...9910...00}$, $k-$ zeros and $n+k+1$ digits. But this sum equals to: $\underbrace{999...9}_\text{n-1} 8 \underbrace{0...0000}_\text{n-1}1$, which has sum of digits $n$. $\blacksquare$