$1 \cdot 60$ Find the smallest positive integer $n>1$, such that the arithmetic mean of $1^{2}, 2^{2}, 3^{2}, \cdots, n^{2}$ is a perfect square.
【Solution】Let
$$
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=k^{2}
$$

where $k \in \mathbb{N}$. That is,
$$
\frac{1}{6}(n+1)(2 n+1)=k^{2}
$$

Since $n \geqslant 2$, we have
$$
\frac{1}{6}(n+1)(2 n+1) \geqslant \frac{5}{2},
$$

thus
$$
\begin{array}{l}
k^{2} \geqslant \frac{5}{2}, \\
k \geqslant 2.
\end{array}
$$

Since $2 n+1$ is odd, it follows from (1) that $n$ must be odd. Let $n=2 m-1, m \geqslant 2$, substituting into (1) gives
$$
\frac{1}{3} m(4 m-1)=k^{2}
$$

Therefore, $m$ is a multiple of 3 or $4 m-1$ is a multiple of 3.
If $m$ is a multiple of 3, then let $m=3 t, t \in \mathbb{N}$, substituting into (2) gives
$$
t(12 t-1)=k^{2},
$$

Since $t$ and $12 t-1$ are coprime, both $t$ and $12 t-1$ must be perfect squares. But
$$
12 t-1 \equiv 3(\bmod 4),
$$

which contradicts the fact that "a perfect square modulo 4 can only be 0 or 1". Therefore, it must be that $4 m-1=3 t (t \in \mathbb{N})$, substituting into (2) gives
$$
\begin{array}{c}
\frac{1}{4}(3 t+1) t=k^{2}, \\
(3 t+1) t=(2 k)^{2}
\end{array}
$$

Since $m=\frac{1}{4}(3 t+1)$, $3 t+1$ is a multiple of 4, thus the remainder of $t$ when divided by 4 must be 1. Let
$$
t=4 j+1 \quad(j \in \mathbb{N})
$$
(because $m \geqslant 2$, hence $t \geqslant 3$, thus $j \geqslant 1$). Substituting into (3), we get
$$
(3 j+1)(4 j+1)=k^{2}.
$$

Since $(4 j+1)-(3 j+1)=j$, and $4 j+1$ and $j$ are coprime, $4 j+1$ and $3 j+1$ are coprime, thus both $4 j+1$ and $3 j+1$ are perfect squares. Let
$$
\left\{\begin{array}{l}
3 j+1=a^{2} \\
4 j+1=b^{2}
\end{array}\right.
$$

where $a$ is a positive integer not less than 2, and $b$ is an odd number not less than 3. From the above, we get
$$
\begin{array}{l}
3 b^{2}+1=4 a^{2}, \\
b=2 s+1, s \in \mathbb{N},
\end{array}
$$

Substituting into the above, we get
$$
\begin{array}{l}
12 s(s+1)+4=4 a^{2}, \\
3 s(s+1)+1=a^{2},
\end{array}
$$

Since $s(s+1)$ is always even, $a$ is odd, and $a \geqslant 3$. Thus, let
$$
b=4 c \pm 1, a=4 d \pm 1,
$$

where $c, d \in \mathbb{N}$. Substituting into (4), we get
$$
3(4 c \pm 1)^{2}+1=4(4 d \pm 1)^{2},
$$

Expanding and simplifying, we have
$$
3 c(2 c \pm 1)=4 d(2 d \pm 1),
$$

From the above, we know that $c$ is a multiple of 4. Thus, let
$$
c=4 e, e \in \mathbb{N},
$$

Therefore,
$$
b=4 c \pm 1=16 e \pm 1,
$$

From
$$
4 j+1=b^{2}
$$

we get

Thus
$$
j=\frac{b^{2}-1}{4}=\frac{(16 e \pm 1)^{2}-1}{4}=8 e(8 e \pm 1),
$$

So
$$
t=4 j+1=32 e(8 e \pm 1)+1,
$$

Thus
$$
m=\frac{1}{4}(3 t+1)=24 e(8 e \pm 1)+1,
$$

Clearly
$$
\begin{array}{l}
n=2 m-1=48 e(8 e \pm 1)+1, \\
n \geqslant 48 \times 1 \times(8 \times 1-1)+1=337.
\end{array}
$$

It is easy to verify that when $n=337$,
$$
\begin{aligned}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n} & =\frac{1}{6}(n+1)(2 n+1) \\
& =\frac{1}{6} \times 338 \times 675=169 \times 225 \\
& =(13 \times 15)^{2}.
\end{aligned}
$$
In summary, the smallest positive integer $n$ greater than 1 is 337.