2. Given the integer $n>2020, \frac{n-2020}{2120-n}$ is a perfect square. Then the value of $n$ is $\qquad$ .
2.2070 or 2100 or 2110.

Let $\frac{n-2020}{2120-n}=m$ (where $m$ is a perfect square).
Then $m=\frac{n-2020}{2120-n}=\frac{100}{2120-n}-1$
$\Rightarrow(m+1)(2120-n)=100$
$\Rightarrow(m, n)=(1,2070)$ or $(4,2100)$ or
$(9,2110)$
$\Rightarrow n=2070$ or 2100 or 2110.