Assume that $n$ satisfies the condition of the problem, and choose arbitrary positive integers $a, b$ satisfying $n! | a^{n}+b^{n}$ and $\frac{a^{n}+b^{n}}{n!} \le 100$. (Case 1) $n$ is even and $n \ge 10$. Since $3|n!|a^{n}+b^{n}$, $a, b$ are both divisible by $3$. Then $$ \frac{a^{n}+b^{n}}{n!} \ge \frac{3^{n}}{3^{v_{3}(n!)}} \ge 3^{\frac{n}{2}} \ge 3^{5} > 100,$$where $v_{3}(m)$ denotes the largest exponenet $e$ satisfying $3^{e}|m$. Contradiction. (Case 2) $n$ is odd and $n \ge 17$. Since $3|n!|a^{n}+b^{n}$, $3|a+b$. Thus $v_{3}(a^{n}+b^{n})=v_{3}(a+b)+v_{3}(n)$. Hence in particular, $a+b \ge 3^{v_{3}(a^{n}+b^{n})-v_{3}(n)} \ge 3^{v_{3}(n!)-v_{3}(n)}$, so by H\"older, $$a^{n}+b^{n} \ge \frac{(a+b)^{n}}{2^{n-1}} \ge \left(\frac{3^{v_{3}(n!)-v_{3}(n)}}{2}\right )^{n}$$Since $n! \le \left (\frac{n}{2}\right )^{n}$, $$\frac{a^{n}+b^{n}}{n!} \ge \left(\frac{3^{v_{3}((n-1)!)}}{n}\right )^{n} \ge \left ( \frac{\sqrt{3}^{n-1-4\log_{3}(n)}}{n}\right )^{n} > 100$$. Contradiction. We are left to check $n=1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15$, which can be done on a case-by-case basis. The answer is $n=1, 2, 3$.