Searching for Perfect Cubes

We certainly want $k^3$ to be an integer, so we want $n-4$ to divide $n^3 - 3n^2 - 4n - 30$ . Since $\frac{n^3 - 3n^2 - 4n - 30}{n-4} \; = \; n^2 + n - \frac{30}{n-4}$ we want $n-4$ to divide $30$ . Thus $n-4$ must be one of $\pm1$ , $\pm2$ , $\pm3$ , $\pm5$ , $\pm6$ , $\pm10$ , $\pm15$ and $\pm30$ . A simple check finds that only two of these 16 options gives a cubic quotient $k^3$ . If $n-4=1$ then $n=5$ , $k=0$ , and if $n-4=2$ then $n=6$ , $k=3$ . Thus the answer is $5+6+2 = \boxed{13}$ .