2. Try to find the largest integer $k$, such that for each positive integer $n$, we have
$$
1980^{k} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}} .\right.
$$
Notice,
$$
\begin{array}{l}
1980=2^{2} \times 3^{2} \times 5 \times 11, \\
v_{11}(1980!)=\sum_{i \geqslant 1}\left[\frac{1980}{11^{i}}\right]=197 . \\
\text { Since }\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right], \text { hence } \\
v_{5}(m!) \geqslant 2 v_{11}(m!) .
\end{array}
$$

Since $\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right]$, hence
similarly, $v_{3}(m!) \geqslant 3 v_{11}(m!)$,
$$
v_{2}(m!) \geqslant 5 v_{11}(m!) \text {. }
$$

Then the maximum $k$ satisfying $1980^{k} \mid 1980!$ is 197.
For any prime $p$,
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \\
=v_{p}((1980 n)!)-1980 v_{p}(n!) \\
=\frac{1980 S_{p}(n)-S_{p}(1980 n)}{p-1} .
\end{array}
$$

Using $S_{p}(m n) \leqslant S_{p}(m) S_{p}(n)$, we get
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \geqslant \frac{1980-S_{p}(1980)}{p-1} S_{p}(n) \\
\geqslant v_{p}(1980!) .
\end{array}
$$

Therefore, for any positive integer $n$, we have
$$
1980^{197} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}}\right. \text {. }
$$

In summary, the maximum $k$ is 197.