Mind Blown!?!?!?

There are $4950$ pairs of points, of which we choose $4026$ to connect with segments. Exactly $1275$ of these pairs have labels differing by at least $50.$ So we'd expect that $1275/4950$ of the segments join such pairs, which would give us an answer of $\frac{1275}{4950} \cdot 4026 = \fbox{1037}.$ This is indeed the correct answer.

Why? In general, if $S$ and $T$ are two subsets of a finite set $X,$ chosen randomly among all the subsets with size $a$ and $b$ respectively, then the expected size of $S \cap T$ is $\frac{ab}{|X|}.$ This is due to the fact that the expected value of a sum of random variables is additive , even if the variables are dependent. We can apply this by writing $T = \{t_1, t_2, \ldots, t_b\},$ and considering the random variables $Y_i = \begin{cases} 1 &\text{if } t_i \in S \\ 0 & \text{if } t_i \notin S. \end{cases}$ Then $E(|S \cap T|) = \sum_i E(Y_i) = b E(Y_1) = b \cdot a/|X|,$ as desired.