Iberoamerican Mathematical Olympiad

Solution:
$0^2 = 0$, $(\pm 1)^2 = 1$, $(\pm 2)^2 = 4$, $(\pm 3)^2 = 0$, $(\pm 4)^2 = 7 \bmod 9$, so the condition is necessary.

We exhibit squares which give these values.

$0 \bmod 9$. Obviously $0^2 = 0$. We have $9^2 = 81$, $99^2 = 9801$ and in general $99\ldots9^2 = (10^n - 1)^2 = 10^{2n} - 2 \cdot 10^n + 1 = 99\ldots980\ldots01$, with digit sum $9n$.

$1 \bmod 9$. Obviously $1^2 = 1$ with digit sum $1$, and $8^2 = 64$ with digit sum $10$. We also have $98^2 = 9604$, $998^2 = 996004$, and in general $99\ldots98^2 = (10^n - 2)^2 = 10^{2n} - 4 \cdot 10^n + 4 = 99\ldots960\ldots04$, with digit sum $9n + 1$.

$4 \bmod 9$. Obviously $2^2 = 4$ with digit sum $4$, and $7^2 = 49$ with digit sum $13$. Also $97^2 = 9409$ with digit sum $22$, $997^2 = 994009$ with digit sum $31$, and in general $99\ldots97^2 = (10^n - 3)^2 = 10^{2n} - 6 \cdot 10^n + 9 = 99\ldots940\ldots09$, with digit sum $9n + 4$.

$7 \bmod 9$. Obviously $4^2 = 16$, with digit sum $7$. Also $95^2 = 9025$, digit sum $16$, $995^2 = 990025$ with digit sum $25$, and in general $99\ldots95^2 = (10^n - 5)^2 = 10^{2n} - 10^{n+1} + 25 = 99\ldots90\ldots025$, with digit sum $9n - 2$.