Example 6 (14th All-Soviet Mathematical Olympiad) Write down the two-digit numbers from 19 to 80 consecutively. Question: Is the number $19202122 \cdots 787980$ divisible by 1980?
Let $A=19202122 \cdots 787980$. Clearly, $20 \mid A$, since
$$
100^{k}=(99+1)^{k} \equiv 1(\bmod 99),
$$

we have
$$
\begin{aligned}
A & =19 \cdot 100^{61}+20 \cdot 100^{60}+21 \cdot 100^{59}+\cdots+78 \cdot 100^{2}+79 \cdot 100+80 \\
& =19+20+21+\cdots+78+79+80=31 \cdot 99 \equiv 0(\bmod 99) .
\end{aligned}
$$

Thus, $99 \mid A$, and since $(20,99)=1$, it follows that $20 \cdot 99 \mid A$, i.e., $1980 \mid A$.