1. Find all positive integers $n$ such that the equation in $x, y$
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{n}
$$

has exactly 2011 solutions in positive integers $(x, y)$ with $x \leqslant y$.
1. From the given, we have
$$
x y-n x-n y=0 \Rightarrow(x-n)(y-n)=n^{2} \text {. }
$$

Therefore, apart from $x=y=2 n$, if $x-n$ takes a positive divisor of $n^{2}$ that is less than $n$, we can obtain a set of positive integer solutions $(x, y)$ that satisfy the condition.

Thus, the number of positive divisors of $n^{2}$ that are less than $n$ is exactly 2010.

Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are distinct prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are non-negative integers. Then the number of positive divisors of $n^{2}$ that are less than $n$ is
$$
\begin{array}{l}
\frac{\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{k}+1\right)-1}{2} . \\
\text { Therefore, }\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{k}+1\right)=4021 .
\end{array}
$$

Since 4021 is a prime number, we have
$$
k=1, \alpha_{1}=2010 \text {. }
$$

Thus, $n=p^{2010}$ (where $p$ is a prime number).