The sum of the first $n$ squares equals $n(n+1)(2 n+1) / 6$, so we require $n(n+1)(2 n+1)$ to be divisible by $600=24 \cdot 25$. The three factors are pairwise relatively prime, so one of them must be divisible by 25 . The smallest $n$ for which this happens is $n=12$ $(2 n+1=25)$, but then we do not have enough factors of 2 . The next smallest is $n=24(n+1=25)$, and this works, so 24 is the answer.