$$
\begin{array}{l}
M=\{(m, n) \mid m \text { and } n \text { are coprime, } m+n>2020, \\
\quad 0<m<n \leqslant 2020\} . \\
\text { Find the sum } S=\sum_{(m, n) \in M} \frac{1}{m n} .
\end{array}
$$
To generalize, replace 2020 with any integer $k \geq 2$. For this, take
$$
M_{k}=\{(m, n) \mid m \text{ and } n \text{ are coprime, } m+n>k,
$$
$$
0k \text{, } \\
0k+1,0 \\
k+1,0k+1,0k+1,0<m<n \leqslant k\} \\
$$
$$
\triangleq M^{\prime \prime} \cup M^{*} \text{. }
$$

Then $s_{k+1}-s_{k}=\sum_{M} \frac{1}{m n}-\sum_{M} \frac{1}{m n}$
$$
=\sum_{m \leqslant k} \frac{1}{m(k+1)}-\sum_{m<\frac{k+1}{2}} \frac{1}{m(k+1-m)} \text{. }
$$

The summation range of the last sum on the right-hand side is due to $m<n=k+1-m$, then $m<\frac{k+1}{2}$.
$$
\begin{array}{l}
\text{Since } \sum_{m<\frac{k+1}{2}} \frac{1}{m(k+1-m)} \\
=\frac{1}{k+1} \sum_{m<\frac{k+1}{2}}\left(\frac{1}{m}-\frac{1}{k+1-m}\right) \\
=\frac{1}{k+1}\left(\sum_{m<\frac{k+1}{2}} \frac{1}{m}+\sum_{m<\frac{k+1}{2}} \frac{1}{k+1-m}\right),
\end{array}
$$

When $1 \leqslant m<\frac{k+1}{2}$, $\frac{k+1}{2}<k+1-m \leqslant k$. Then
$$
\sum_{m<\frac{k+1}{2}} \frac{1}{k+1-m}=\sum_{\frac{k+1}{2}<i \leqslant k} \frac{1}{t}=\sum_{\frac{k+1}{2}<m \leqslant k} \frac{1}{m} \text{. }
$$

And by $m+n=k+1$ and $(m, n)=1$, we know that $\frac{k+1}{2}$ will not be taken by $m$ or $n$.
Thus, the right-hand side of equation (2) can be combined into $\frac{1}{k+1} \sum_{m \leqslant k} \frac{1}{m}$. According to equation (1), we have
$s_{k+1}-s_{k}=\sum_{m \leqslant k} \frac{1}{m(k+1)}-\sum_{m \leqslant k} \frac{1}{m(k+1)}=0$.
Hence, $s_{k}=s_{k-1}=\cdots=s_{2}=\frac{1}{2}$.
Therefore, $s=s_{2020}=\frac{1}{2}$.