Six. (15 points) Find all positive integers $n$, such that there exist non-zero integers $x_{1}, x_{2}, \cdots, x_{n}, y$, satisfying
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=0, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n y^{2} .
\end{array}\right.
$$
(Xiong Bin provided)
Obviously, $n \neq 1$.
When $n=2 k$ is even, let
$$
x_{2 i-1}=1, x_{2 i}=-1, i=1,2, \cdots, k, y=1 \text {, }
$$

then the conditions are satisfied.
When $n=3+2 k\left(k \in \mathbf{N}_{+}\right)$, let
$$
\begin{array}{l}
y=2, x_{1}=4, x_{2}=x_{3}=x_{4}=x_{5}=-1, \\
x_{2 i}=2, x_{2 i+1}=-2, i=3,4, \cdots, k+1,
\end{array}
$$

then the conditions are satisfied.
When $n=3$, if there exist non-zero integers $x_{1}, x_{2}, x_{3}$ such that $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=0, \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=3 y^{2},\end{array}\right.$ then
$$
2\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}\right)=3 y^{2} \text {. }
$$

Assume without loss of generality that $\left(x_{1}, x_{2}\right)=1$, then $x_{1}, x_{2}$ are both odd or one is odd and the other is even. Thus, $x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}$ is odd. On the other hand, $2 \mid y$, so $3 y^{2} \equiv 0(\bmod 4)$. However,
$$
2\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}\right) \equiv 2(\bmod 4),
$$

which is a contradiction.
In summary, the positive integers $n$ that satisfy the conditions are all positive integers except 1 and 3.