Solution. Let $\frac{2020}{2021}=\frac{n}{n+1} \cdot \frac{m}{m+1}$, where $m, n \in \mathbb{N}$. From this, we sequentially obtain

$$
\begin{aligned}
& 2020(n+1)(m+1)=2021 m n \\
& 2020(m+n+1)=m n \\
& n=\frac{2020(m+1)}{m-2020} \\
& n=\frac{2020(m-2020)+2020 \cdot 2021}{m-2020} \\
& n=2020+\frac{2020 \cdot 2021}{m-2020}
\end{aligned}
$$

Since $m, n \in \mathbb{N}$, $m-2020$ must be a positive divisor of $2020 \cdot 2021$, and each divisor of $2020 \cdot 2021$ corresponds to exactly one pair $(n, m)$. We have

$$
2020 \cdot 2021=2^{2} \cdot 5 \cdot 43 \cdot 47 \cdot 101
$$

so the number of divisors of $2020 \cdot 2021$ is equal to $3 \cdot 2^{4}=48$. Finally, there are 48 ways to write the fraction $\frac{2020}{2021}$ as a product of two fractions of the desired form.
2. In how many ways can the fraction $\frac{2020}{2021}$ be written as a product of two fractions of the form $\frac{n}{n+1}, n \in \mathbb{N}$.