2. From $1,2, \cdots, 100$ choose three different numbers such that they cannot form the three sides of a triangle. The number of different ways to do this is.
2. 82075.

Let these three numbers be $i, j, k(1 \leqslant i<j<k \leqslant 100)$. Then $k \geqslant i+j$.
Therefore, the number of different ways to choose these three numbers is
$$
\begin{array}{l}
\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100}\left(\sum_{k=i+j}^{100} 1\right)\right]=\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100-i}(101-i-j)\right] \\
=\sum_{i=1}^{100}\left(\sum_{i=1}^{100-2 i} t\right)=\sum_{i=1}^{49} \frac{(100-2 i)(101-2 i)}{2} \\
=\sum_{k=1}^{49} k(2 k+1)(k=50-i) \\
=\frac{49 \times 50 \times(2 \times 49+1)}{3}+\frac{49 \times 50}{2} \\
=82075 .
\end{array}
$$