What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square? 
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?

Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.
From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2}
$$
is this of any help here?
Remark.  Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2

The mean of the squares $1^2, \ldots, n^2$ is
$$ f(n) = \frac{1}{n} \sum_{i=1}^n i^2 = \frac{2n^2+3n+1}{6}$$
It is an integer if and only if $n \equiv 1$ or $5 \mod 6$.
The first $n > 1$ for which it is a square is $337$, where $f(337) = 38025 = 195^2$.  There are infinitely many. See OEIS sequence A084231.
