2. Let the sum of the squares of the first 101 positive integers starting from a positive integer $k$ be equal to the sum of the squares of the next 100 positive integers. Then the value of $k$ is $\qquad$ .
2.20100.

From the problem, we get
$$
k^{2}+\sum_{i=1}^{100}(k+i)^{2}=\sum_{i=1}^{100}(k+100+i)^{2} \text {. }
$$

Then $k^{2}=\sum_{i=1}^{100}\left[(k+100+i)^{2}-(k+i)^{2}\right]$
$$
=200 \sum_{i=1}^{100}(k+50+i)=10000(2 k+201) \text {. }
$$

Thus, $k^{2}-20000 k-2010000=0$.
Taking the positive root, we get $k=20100$.