Example 10 (1980 14th All-Soviet Union Mathematical Olympiad) Write down the two-digit numbers from 19 to 80 consecutively. Is the resulting number $19202122 \cdots 787980$ divisible by 1980?
Let $A=19202122 \cdots 7980$.
Obviously, $20 \mid A$.
Since $100^{k}=(99+1)^{k}=99 M+1$, where $M$ is a positive integer and $k$ is a positive integer,
we have $100^{k} \equiv 1(\bmod 99)$.
Therefore,
$$
\begin{aligned}
A & =19 \cdot 100^{61}+20 \cdot 100^{60}+\cdots+79 \cdot 100+80 \\
& \equiv(19+20+21+\cdots+79+80) \\
& \equiv 31 \cdot 99 \\
& \equiv 0(\bmod 99),
\end{aligned}
$$

Thus, $99 \mid A$.
Since $(20,99)=1$, we have
$$
20 \cdot 99=1980 \mid A.
$$