jmc

Notice that $n^3-n$ factors as $n^3-n=n(n^2-1)=(n-1)n(n+1)$. We observe that among any three consecutive integers, at least one must be divisible by $2$ and one must be divisible by $3$. Thus, we know that $6$ must always divide into $n^3-n$. Indeed, this is the largest such integer; for $n=6$, then $n^3-n=210=6\cdot 5\cdot 7$, and for $n=33$, then $n^3-n=32\cdot 33\cdot 34=6\cdot 32\cdot 11\cdot 17$, whose greatest common divisor is $\boxed{6}$.