$n\ge 2$.

Rust wrote: $n\ge 2$. Why? Post your solution!

Use the RMS-AM inequality.....

For any permutation $\sigma$, if $\{x_i\}-$ solution, then $\{x_{\sigma(i)}\}$ solution too. Therefore we can find only solutions $0\le x_1\le x_2\le ....\le x_n$. Let $s=\sum_i x_i, y_i=nx_i-s \to \sum_i y_i=0$. Then $\sum_i (y_i+s)^2=n^2+\frac{4n^2}{4n+1}s^2$ or $\sum_ i y_i^2=n(n-\frac{s^2}{4n+1}).$ Therefore $(4n+1)|s^2$ and $s^2\le n(4n+1)$. Let $P=\prod_{p|4n+1, v_p(4n+1)-odd} p$. Then $4n+1=Pq^2$ and $s=Pqk$. Then $\sum_i y_i^2=n(n-Pk^2), Pk^2\le n, \sum y_i=0$ give solution, if $x_i=\frac{s+y_i}{n}\in Z$. $n=1, 4n+1=5\not =pq^2$ - non solution $n=2, 4n+1=9=1*3^2, s=3\to x_1=1,x_2=2$ - have solution. $n=3,4,5$ - have not solution. $n=6, 4n+1=5^2\to s=5k$ have solution $y_1=-5, y_2=...=y_6=1$ or $x_1=0,x_2=...=x_6=1$. We can find for all $n=2+9k$, when $9|4n+1$ by add $x_i=0$ 6 times and $x_i=4$ 3 times.

Sorry,I was wrong, it won't work with that.