9. (40 points) Find the number of pairs of natural numbers $m$ and $n$, satisfying the equation $\frac{1}{m}+\frac{1}{n}=\frac{1}{2020}$.
Answer: 45.

Solution: Transform the given equation

\[
\begin{gathered}
\frac{1}{m}+\frac{1}{n}=\frac{1}{2020} \Leftrightarrow \frac{n+m}{m \cdot n}=\frac{1}{2020} \Leftrightarrow m n=2020(n+m) \Leftrightarrow \\
\Leftrightarrow m n-2020 n-2020 m+2020^{2}-2020^{2}=0 \Leftrightarrow(n-2020)(m-2020)=2020^{2}
\end{gathered}
\]

The solutions to this equation are \( m = d + 2020 \) and \( n = \frac{2020^{2}}{d} + 2020 \), where \( d \) is any divisor of \( 2020^{2} \). It remains to find the number of such divisors. Note that \( 2020^{2} = 2^{4} \cdot 5^{2} \cdot 101^{2} \). If \( d \) is a divisor of \( 2020^{2} \), then 2 can appear in its prime factorization in powers \( 0, \ldots, 4 \), and 5 and 101 can appear in powers \( 0, 1, 2 \). Therefore, the total number of such factorizations is \( 5 \cdot 3 \cdot 3 = 45 \).