Answer: A

Small Hint:

Factor the expression as $n^2(n-1)(n+1)$

Big Hint:

Among three consecutive integers there is a multiple of $3,$ and the factors supply at least two powers of $2$

Solution:

We have $$n^2(n^2-1)=n^2(n-1)(n+1).$$ Among $n-1,$ $n,$ and $n+1,$ one is divisible by $3.$ If $n$ is even, $n^2$ is divisible by $4;$ if $n$ is odd, both $n-1$ and $n+1$ are even, so their product is divisible by $4.$ Thus the expression is always divisible by $12.$ It is not always divisible by $24,$ since $n=2$ gives $12.$

Therefore, the correct answer is A.