2. From the 99 natural numbers $1,2,3, \cdots, 99$, the number of ways to choose two different numbers such that their sum is less than 99 is $\qquad$ ways.
2.2352 .

Let the two selected numbers be $x$ and $y$, then $x+y$ has three cases:
$$
x+y=100, x+y<100, x+y>100 \text {. }
$$

Below, we only consider the first two scenarios:
(1) When $x+y=100$, the number of ways to select is 49;
(2) When $x+y<100$. This indicates that the number of ways to select when $x+y<100$ is equal to the number of ways when $x+y>100$, and their values are both $\frac{\mathrm{C}_{49}^{2}-49}{2}$.

In the case of $x+y<100$, there are $x+y=99$ and $x+y<99$, and the selection method for $x+y=99$ should be excluded, which is 49.
Therefore, the number of ways that satisfy the condition is $\frac{\mathrm{C}_{99}^{2}-49}{2}-49=2352$.