这是一篇劝退版的题解。

$$\begin{aligned} \frac{1}{x}+\frac{1}{y}=\frac{1}{n!} \text{的正整数解的个数} &= \sum_{x \in N^*} \sum_{y \in N^*} [\frac{1}{x}+\frac{1}{y}=\frac{1}{n!}] \\ &= \sum_{x \in N^*} \sum_{y \in N^*} [\frac{xy}{x+y}=n!] \\ &= \sum_{x \in N^*} \sum_{y \in N^*} [xy=n!x+n!y] \\ &= \sum_{x \in N^*} \sum_{y \in N^*} [xy-n!y=n!x] \\ &= \sum_{x \in N^*} \sum_{y \in N^*} [y(x-n!)=n!x] \\ &= \sum_{x \in N^*} \sum_{y \in N^*} [y=\frac{n!x}{x-n!}] \\ &= \sum_{x \in N^*} [\frac{n!x}{x-n!} \in N^*] \\ &= \sum_{n!+1} ^{(n!)^2+n!} [\frac{n!x}{x-n!} \in N^*] \ \ \ \ \text{这一步跳了若干步确定上下界的步骤} \\ &= \sum_{1} ^{(n!)^2} [\frac{(n!)^2 + k \times n!}{k} \in N^*] \ \ \ \ \ \text{换元，令} k=x-n! \\ &= \sum_{1} ^{(n!)^2} [k \mid ((n!)^2 + k \times n!)] \\ &= \sum_{1} ^{(n!)^2} [k \mid (n!)^2] \\ &= \operatorname{d}((n!)^2) \text{注：}\operatorname{d}(n) \text{为 n 的约数个数} \end{aligned} $$