$6 \cdot 103$ 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 defined as
$$
\left(\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right)^{\frac{1}{2}} .
$$
\[ \text{Solution: Let } \left(\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}\right)^{\frac{1}{2}}=m, m \in \mathbb{N}. \]
Then
\[ \begin{array}{l}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2}, \\
\frac{(n+1)(2 n+1)}{6}=m^{2}, \\
(n+1)(2 n+1)=6 m^{2}.
\end{array} \]

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 \(6 p+3, p \in \mathbb{N} \cup \{0\}\).
\[ \begin{array}{c}
\text{If } n=6 p+1, \text{ then} \\
6 m^{2}=(6 p+2)(12 p+3), \\
m^{2}=(3 p+1)(4 p+1).
\end{array} \]

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 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 perfect square 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. Therefore, \(n\) 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. Therefore, \(n\) cannot be of the form \(6 p+3\).
In conclusion, the smallest positive integer \(n\) is 337.