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 $.