All integers of the form $499\ldots99$ satisfy the condition. Indeed, if $n = 4\underbrace{99\ldots99}_{k} = 5 \cdot 10^k - 1$ then
$$n^2 = 25 \cdot 10^{2k} - 10^{k+1} + 1 = 24 \underbrace{99\ldots9}_{k-1} \underbrace{00\ldots00}_{k} 1.$$
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.