Test $\mathbf{A}$ Find the smallest positive integer $n$ (where $n>1$) such that the average of the squares of the first $n$ natural numbers is a perfect square.
Let
\[
\begin{aligned}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n} & =m^{2}(m \in \mathbf{N}) \Rightarrow \frac{1}{6}(n+1)(2 n+1) \\
& =m^{2} \Rightarrow(n+1)(2 n+1)=6 m^{2} .
\end{aligned}
\]

Since \(6 m^{2}\) is even and \(2 n+1\) is odd, \(n+1\) must be even, hence \(n\) is odd.
We classify \(n\) modulo 6, setting \(n=6 p \pm 1\) or \(n=6 p+3\).
(1) When \(n=6 p+3\)
\[
6 m^{2}=(6 p+4)(12 p+7)=72 p^{2}+90 p+28,
\]

Since \(6 \mid 72\), \(6 \mid 90\), and \(6 \times 28\), there is no solution in this case.
(2) When \(n=6 p-1\)
\[
6 m^{2}=6 p(12 p-1) .
\]

Since \((p, 12 p-1)=1\), for the equation to hold, both \(p\) and \(12 p-1\) must be perfect squares. Let \(p=s^{2}\) and \(12 p-1=t^{2}\), then \(t^{2}=12 s^{2}-1\). Since perfect squares can only be of the form \(4 k\) or \(4 k+1\), \(12 s^{2}-1=4(3 s^{2})-1\) is not a perfect square. Hence, there is no solution in this case.
(3) When \(n=6 p+1\)
\[
6 m^{2}=(6 p+2)(12 p+3) \Rightarrow m^{2}=(3 p+1)(4 p+1) .
\]

Since \((3 p+1, 4 p+1)=1\), both \(3 p+1\) and \(4 p+1\) must be perfect squares. Let \(3 p+1=u^{2}\) and \(4 p+1=v^{2}\), then \(4 u^{2}-3 v^{2}=1\). Clearly, \(u=v=1\) is one solution, in which case \(p=0\) and \(n=1\), contradicting \(n>1\). Testing \(u=2,3,4, \cdots, 11, 12\), \(v\) is not an integer. When \(u=13\), \(v=15\) is a solution, giving \(p=56\), \(m=195\), and \(n=337\). Therefore, the smallest positive integer \(n\) is 337.

The issue here is that \(n\) is found through trial and error, which would significantly increase the number of trials for the next \(n\). Although other analytical methods can be used to find \(n\), they do not make finding the next \(n\) much easier. Another approach is:

From \(4 u^{2}-3 v^{2}=1\), we know \(v\) is odd. Let \(v=2 q+1\), then the equation becomes \(u^{2}-3 q(q+1)-1=0\).

Since \(q(q+1)\) is even, \(u\) must be odd. Let \(u=2 j+1\), then the equation becomes \(4 j(j+1)=3 q(q+1)\). The left side is a multiple of 8. To make the right side a multiple of 8 and find the smallest \(n\), we can set \(q+1=8\), giving \(q=7\), \(j=6\), and \(j+1=7\). Thus, \(u=2 j+1=13\) and \(v=2 q+1=15\), so \(n=337\).