All Even

Note that $n(n^2-1) = (n-1)n(n+1)$ is the product of three consecutive integers. Amrong three consecutive integers one must be divisible by 3. Therefore, $3 \mid n(n^2-1)$ . Now we are given that $\dfrac n2$ is even. Let $\dfrac n2 = 2m$ , where $m \in \mathbb Z$ . Then $n=4m$ . Therefore $4 \mid n(n^2-1)$ and hence $\boxed{12} \mid n(n^2-1)$ .

$\dfrac{n}{2}$ is even, $\therefore$ $n$ is of the form $4m$ . Then

$(n)(n^{2}-1)$ can be written as $(4m)(4m-1)(4m+1)$ .

Now we get the number $4m$ can be of the form $3k, 3k+1,3k+2$ .

Now in any case we get the number is divisible by $4$ and simultaneously divisible by $3$ .

Hence it is assured that the number is atleast divisible by $\large 3 \times 4=12$