Berkeley Math Circle

Solution:
The number is even, because either $n$ or $n+1$ is even.
Now we show it is always divisible by three. Assume for contradiction that it isn't. Then neither $n$ nor $n+1$ is divisible by three, so $n+2$ must be. However, $2 n+1 = 2(n+2) - 3$ is then also a multiple of three, which is a contradiction.

In fact, one can also notice the result from the fact that
$$1^{2} + 2^{2} + \cdots + n^{2} = \frac{n(n+1)(2 n+1)}{6} .$$