9. (14 points) For any positive integer $n$, let $S(n)$ denote the number of positive integer pairs $(x, y)$ that satisfy the indeterminate equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$. (For example, the positive integer pairs that satisfy $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ are $(6,3),(4,4),(3,6)$, so $S(2)=3$). Find all positive integers $n$ such that $S(n)=2007$.
From $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}\left(x, y, n \in \mathbf{Z}_{+}\right)$, we know $x>n, y>n$.
Let $x=n+a, y=n+b\left(a, b \in \mathbf{Z}_{+}\right)$. Then $\frac{1}{n+a}+\frac{1}{n+b}=\frac{1}{n} \Leftrightarrow n^{2}=a b$.
Therefore, $S(n)$ equals the number of positive integer pairs $(a, b)$. Thus, $S(n)$ equals the number of positive divisors of $n^{2}$.

Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are different primes, and $\alpha_{i} \in \mathbf{Z}_{+}(1 \leqslant i \leqslant k)$. Then
$$
\begin{array}{l}
n^{2}=p_{1}^{2 \alpha_{1}} p_{2}^{2 \alpha_{2}} \cdots p_{k}^{2 \alpha_{k}}. \\
The number of positive divisors of n^{2} is \left(2 \alpha_{1}+1\right) \cdots\left(2 \alpha_{k}+1\right). \\
Let \left(2 \alpha_{1}+1\right) \cdots\left(2 \alpha_{k}+1\right)=2007=3^{2} \times 223. \\
Then \left\{\begin{array}{l}
k=1, \\
\alpha_{1}=1003
\end{array} \text { or } \left\{\begin{array}{l}
k=2, \\
\alpha_{1}=1, \\
\alpha_{2}=334
\end{array}\right.\right. \\
\text { or }\left\{\begin{array}{l}
k=2, \\
\alpha_{1}=4, \\
\alpha_{2}=111
\end{array} \text { or } \left\{\begin{array}{l}
k=3, \\
\alpha_{1}=\alpha_{2}=1, \\
\alpha_{3}=111 .
\end{array}\right.\right. \\
\end{array}
$$

Therefore, the $n$ that satisfies the condition is $n=p_{1}^{1003}$ or $n=p_{1} p_{2}^{334}$ or $n=p_{1}^{4} p_{2}^{111}$ or $n=p_{1} p_{2} p_{3}^{111}$.