Answer: 4

Solution:

Since $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{2} - 1 = \frac{(n+1)(n-2)}{2},$ $$\begin{aligned} \binom{\binom{n}{2}}{2} \\ &= \frac{1}{2} \cdot \frac{n(n-1)}{2} \\ &\quad {}\cdot \frac{(n+1)(n-2)}{2} \\ &= \small \frac{(n+1)n(n-1)(n-2)}{8} \\ &= 3\binom{n+1}{4}. \end{aligned}$$

By the hockey stick identity, $$\begin{aligned} \sum_{n=3}^{40} 3\binom{n+1}{4} &= 3\sum_{k=4}^{41}\binom{k}{4} \\ &= 3\binom{42}{5} \\ &= 3 \cdot 850668 \\ &= 2552004. \end{aligned}$$

The remainder upon division by $1000$ is $4.$