14.5.13 ** Find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.
Let $\frac{1}{n}\left(1^{2}+2^{2}+\cdots+n^{2}\right)$ be a square number $m^{2}$, then $6 m^{2}-(n+1)(2 n+1)$. Since $(n+1,2 n+1)=(n+1,1)=1$, and $2 n+1$ is odd, we have $\left\{\begin{array}{l}n+1=2 u^{2}, \\ 2 n+1=3 v^{2}\end{array}\right.$ or $\left\{\begin{array}{l}n+1=6 u^{2}, \\ 2 n+1=v^{2} .\end{array}\right.$ The latter gives $12 u^{2}-1=v^{2}$. Thus, $v^{2} \equiv-1(\bmod 4)$, which is impossible. From the former, we get $4 u^{2}-3 v^{2}=1$. By testing $u$ from 1 to 13, we find the solutions $u=1, v=1$ (which does not meet the problem's requirements) and $u=13, v=$ 15. This yields the smallest positive integer $n=337$.