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
$$\begin{cases} x_1 + x_2 + x_3 = 0, \\ x_1^2 + x_2^2 + x_3^2 = 3y^2. \end{cases}$$
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 $3y^2 \equiv 0 \pmod 4$. This is a contradiction since $2(x_1^2 + x_2^2 + x_3 x_2) \equiv 2 \pmod 4$.

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