$6 \cdot 103$ Find the smallest positive integer $n$ (where $n>1$) such that the quadratic mean of the first $n$ natural numbers is an integer. Here, the quadratic mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$ is given by
$$\left(\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right)^{\frac{1}{2}} .$$
［Solution］Let $\left(\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}\right)^{\frac{1}{2}}=m, m \in N$.
Then $\square$
$$\begin{array}{l}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2} \\
\frac{(n+1)(2 n+1)}{6}=m^{2} \\
(n+1)(2 n+1)=6 m^{2}
\end{array}$$

Since $6 m^{2}$ is even and $2 n+1$ is odd, $n+1$ must be even, hence $n$ is odd. Therefore, we set $n=6 p \pm 1$ or $6 p+3, p \in N \cup\{0\}$.

If $n=6 p+1$, 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. We set
$$3 p+1=u^{2}, 4 p+1=v^{2}$$

Thus, $4 u^{2}-3 v^{2}=1$
When $u=1$, $v=1, p=0, n=1$, which contradicts $n>1$.
When $u=2,3,4, \cdots, 11,12$, it is easy to verify through calculation that $v$ is not an integer, leading to a contradiction.

When $u=13$, $v=15, p=56, n=337$.
If $n=6 p-1$, 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. We set
$$p=S^{2}, 12 p-1=t^{2}$$

Thus, $t^{2}=12 S^{2}-1$
Since a square number can only be a multiple of 4 or 1 more than a multiple of 4, we have
$$t^{2}=4\left[\left(3 S^{2}\right)-1\right]+3$$

which is impossible. This means that $n$ cannot be of the form $6 p-1$.
$$\begin{array}{l}
\text { If } n=6 p+3 \text {, then } \\
6 m^{2}=(6 p+4)(12 p+7), \\
3 m^{2}=(3 p+2)(12 p+7) \\
3 m^{2}=3 p(12 p+7)+24 p+14 .
\end{array}$$

Since $3 p(12 p+7)$, $24 p$, and $3 m^{2}$ are all multiples of 3, but 14 is not a multiple of 3, the above equation cannot hold. This means that $n$ cannot be of the form $6 p+3$.
In conclusion, the smallest positive integer $n$ is 337.