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All integers of the form $499\dots 99$ satisfy the condition. Indeed, if $n=4\underset{k}{\underbrace{99\dots 99}}=5\cdot 10^k-1$ then $$n^2=25\cdot 10^{2k}-10^{k+1}+1=24\underset{k-1}{\underbrace{99\dots 9}}\underset{k}{\underbrace{00\dots 00}}1.$$ In such a case $s(n)=4+9k$, but $s(n^2)=7+9(k-1)=9k-2$.

Solution: Consider a sequence $10^{3m}-10^{2m}-1$. Similarly to the original solution it is easy to check that if $m$ increases by $1$ then $s(n)$ increases by $27$, but $s(n^2)$ increases by $18$ only.