The sum of the squares of the first $n$ natural numbers is divisible by $n$ when?
I. solution: It is known that

$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$

Since $n, n+1$ and $2 n+1$ are relatively prime numbers, our expression is clearly divisible by $n$ if and only if $\frac{(n+1)(2 n+1)}{6}$ is an integer, that is, $(n+1)(2 n+1)$ is divisible by 2 and 3. Since $(2 n+1)$ is odd, $(n+1)$ must be even, so $n$ is odd. If $n$ is divisible by 3, then neither factor is divisible by 3. Therefore, $n$ can only be of the form $3 k \pm 1$. For $n=3 k+1$, $2 n+1=6 k+2+1=6 k+3$ is divisible by 3, and for $n=3 k-1$, $n+1=3 k$ is divisible by 3.

Thus, the divisibility holds for every $n$ that is not divisible by 2 and 3.

II. solution: Since the left side of the identity in the previous solution is an integer, the right side $\frac{n(n+1)(2 n+1)}{6}$ must also be an integer for every integer $n$. Therefore, if $n$ and 6 are relatively prime, then $\frac{(n+1)(2 n+1)}{6}$ must necessarily be an integer, and thus the sum of squares is divisible by $n$. However, if $n$ and 6 have a common divisor, this cannot be a divisor of either $n+1$ or $2 n+1$, because these are relatively prime to $n$. Thus, in this case, $\frac{(n+1)(2 n+1)}{6}$ cannot be an integer.

Therefore, the divisibility holds for numbers of the form $n=6 k \pm 1$.