[Example 3.5.5] Find the smallest positive integer $n$, such that
$$
\sqrt{\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}}
$$

is an integer, where $n>1$.
To make $\sqrt{\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}}$ an integer, it is necessary for $\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}$ to be a perfect square. For this, we set
$$
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=\frac{(n+1)(2 n+1)}{6}=m^{2} \quad(m \in \mathbf{N}),
$$

then
$$
6 m^{2}=(n+1)(2 n+1)
$$

Since $6 m^{2}$ is even and $2 n+1$ is odd, $n+1$ must be even, and $n$ must be odd. Also, since $(n+1)(2 n+1)$ is a multiple of 6, we should classify $n$ modulo 6. Considering $n$ is odd, we only need to consider $n=6 p+3, 6 p-1$, and $6 p+1$:
(1) If $n=6 p+3(p \in \mathbf{N})$, then
$$
6 m^{2}=(6 p+4)(12 p+7)=72 p^{2}+90 p+28 .
$$

Since 72 and 90 are multiples of 6, but 28 is not, the above equation cannot hold.
There is no solution when $n=6 p+3$.
(2) If $n=6 p-1\left(p \in \mathbf{N}^{*}\right)$, then
$$
\begin{array}{l}
6 m^{2}=6 p(12 p-1), \\
m^{2}=p(12 p-1) .
\end{array}
$$

Since $p$ and $12 p-1$ are coprime, both $p$ and $12 p-1$ must be perfect squares.
However, since a perfect square modulo 4 is 0 or 1, and $12 p-1$ modulo 4 is 3, $12 p-1$ cannot be a perfect square. Thus, there is no solution in this case.
(3) If $n=6 p+1(n \in \mathbf{N})$, then
$$
\begin{array}{l}
6 m^{2}=(6 p+2)(12 p+3), \\
m^{2}=(3 p+1)(4 p+1) .
\end{array}
$$

Since $3 p+1$ and $4 p+1$ are coprime, both $3 p+1$ and $4 p+1$ must be perfect squares. Let
$$
\left\{\begin{array}{l}
3 p+1=u^{2} \\
4 p+1=v^{2}
\end{array}\right.
$$

Eliminating $p$ gives
$$
3 v^{2}+1=4 u^{2} .
$$

Clearly, $u=v=1$ is a solution, but this gives $p=0, n=1$, which contradicts the condition $n>1$.
We now check for $u=2,3,4, \cdots$.
When $u=2,3, \cdots, 12$, $v$ is not an integer.
When $u=13$, $v=15$, at this point $p=56, m=195, n=337$.
Thus, the smallest natural number $n$ greater than 1 is $n=337$.