Note that $0 \leq r(n, 1000) \leq 999$ and $0 \leq r(n, 1001) \leq 1000$. Consider the $\binom{1000}{2}=499500$ ways to choose pairs $(i, j)$ such that $i>j$. By the Chinese Remainder Theorem, there is exactly one $n$ such that $1 \leq n \leq 1000 \cdot 1001$ such that $n \equiv i(\bmod 1000)$ and $n \equiv j(\bmod 1001)$. Finally, it is easy to check that none of the $n$ in the range 1000001 to 1001000 satisfy the condition, so the answer is exactly 499500.