The conditions of the problem can be written the following way: $\frac{n}{2020} < \frac{1}{100} < \frac{n+1}{2019}$.

The left inequality implies that $\frac{n}{2020} < \frac{1}{100}$, so $n < \frac{2020}{100} = 20.2$, hence $n \le 20$.

If $n=20$ is substituted into the right inequality, one can see that it holds: $\frac{21}{2019} > \frac{1}{100} \Leftrightarrow 2100 > 2019$.

The right inequality doesn't hold for $n \le 19$, since $\frac{n+1}{2019} \le \frac{20}{2019} < \frac{20}{2000} = \frac{1}{100}$.

Thus, the only possible solution is $n=20$.