The answer is $n\cdot \frac{\binom{d}{2}}{3}=2000$ Take the obvious graph interpretation. $G$ is $d$-regular and has $n$ vertices, $(d,n=4,1000)$. Each vertex can be in at most $\binom{d}{2}$ triangles for obvious reasons, so the number of triangles is at most $n\cdot \frac{\binom{d}{2}}{3}$. Equality occurs with $K_{d+1}$ copied $\frac{n}{d+1}$ times.