Three, find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.
(Note: The square mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{i}$ is defined as $\left.\left[\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right]^{\frac{1}{2}}\right)$
$$
\begin{array}{c}
\text { Three, let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \\
=\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \text { be a }
\end{array}
$$

perfect square $m^{2}$, then $6 m^{2}=(n+1)(2 n+1)$. Since $6 m^{2}$ is even, it must be odd. We can set $n=3 p+q$, where $q=-1, 1$ or 3.

If $q=2$, then $6 m^{2}=(6 p+4)(12 p+7)$ $=72 p^{2}+90 p+28$. Since $6|72, 6| 90$, but $6 \nmid 28$, this case is impossible.

If $q=-1$, then $6 m^{2}=(6 p)(12 p-1)$, i.e., $m^{2}=p(12 p-1)$. Since $(p, 12 p-1) = 1$, both $p$ and $12 p-1$ must be perfect squares. Let $p=s^{2}$, $12 p-1=t^{2}$, we get $t^{2}-12 s^{2}=-1$. Since $t^{2} \equiv 0$ (mod 3), this case is also impossible.

If $q=1$, then $6 m^{2}=(6 p+2)(12 p+3)$, i.e., $m^{2}=(3 p+1)(4 p+1)$. Since $(3 p+1, 4 p+1)=1$, both $3 p+1=u^{2}$ and $4 p+1=v^{2}$, we get $3 v^{2}+1=4 u^{2}$. Clearly, $u=v=1$ is a solution, but it leads to $p=0, m=1, n=1$, which does not meet the requirement. To find a solution for $n>1$, by trying integer values of $u$ from 2 to 12, $v$ has no integer value, but when $u=13$, $v=15$. Therefore, the smallest solution is $p=56, m=195, n=337$. The answer is $n=337$.