Solution:

Note that, for any real numbers $a$ and $b$, the segment connecting $(-1, a)$ and $(1, b)$ has midpoint $\left(0, \frac{a+b}{2}\right)$, so its $y$-intercept is $\frac{a+b}{2}$. Now suppose that our given segments connect $(-1,1)$ to $\left(1, y_{1}\right)$, $(-1,2)$ to $\left(1, y_{2}\right)$, $(-1,3)$ to $\left(1, y_{3}\right)$, $\ldots$, $(-1,2000)$ to $\left(1, y_{2000}\right)$. The sum of the $y$-intercepts is then

$\frac{1+y_{1}}{2}+\frac{2+y_{2}}{2}+\frac{3+y_{3}}{2}+\cdots+\frac{2000+y_{2000}}{2}=\frac{1}{2}\left[(1+2+3+\cdots+2000)+\left(y_{1}+y_{2}+y_{3}+\cdots+y_{2000}\right)\right]$.

But it follows from the given that $y_{1}, y_{2}, \ldots, y_{2000}$ are just some reordering of $1,2, \ldots, 2000$, so our sum simplifies to
$$\frac{1}{2}[(1+2+3+\cdots+2000)+(1+2+3+\cdots+2000)]=1+2+3+\cdots+2000$$
By the formula for the sum of an arithmetic progression, this equals $2000 \cdot 2001 / 2 = 2,001,000$.