jmc

To have three distinct elements, two of $n-2$, $n+2$, $2n$, and $\frac{n}{2}$ must be equal. It is clear that $n-2$ can never equal $n+2$. However, any other equality pairing among the four is possible, so we simply check for distinctness. If $2n=\frac{n}{2}$, the solution is $n=0$. If $n+2=2n$, $n=2$. If $n-2=2n$, $n=-2$. If $\frac{n}{2}=n-2$, $n=4$. Finally, if $\frac{n}{2}=n+2$, $n=-4$. Thus, there are $\boxed{5}$ such $n$.