China Western Mathematical Olympiad

It is easy to see that $n>1$.

When $n=2k$, $k\in \mathbb{N}$, let $x_{2i-1}=1$, $x_{2i}=-1$, $i=1,2,\dots ,k$, and $y=1$, then the condition is satisfied.

When $n=2k+3$, $k\in \mathbb{N}$, let $y=2$, $x_1=4$, $x_2=x_3=x_4=x_5=-1$, $x_{2i}=2$, $x_{2i+1}=-2$, $i=3,4,\cdots ,k+1$, then the condition is satisfied.

Now if $n=3$, and there exist $x_1,x_2,x_3,y$ such that $$\{\begin{array}{l}{x}_1+x_2+x_3=0,\\ x_1^2+x_2^2+x_3^2=3y^2.\end{array}$$ then $$2(x_1^2+x_2^2+x_3{x}_2)=3y^2.$$ Suppose $\gcd (x_1,x_2)=1$, so $x_1,x_2$ are odd numbers or one is even while the other is odd. Hence $x_1^2+x_2^2+x_3{x}_2$ is an odd number. But $2\mid 3y^2$, so $3y^2\equiv 0(\mathrm{mod}4)$. This is a contradiction since $2(x_1^2+x_2^2+x_3{x}_2)\equiv 2(\mathrm{mod}4)$.

The answer is $n\ne 1,3$.