By # 4216, we have $9^{50}\equiv 1\pmod{1000}$. Therefore, we have $$9\equiv 9^{51}\equiv 9^{101}\equiv\cdots, 9^2\equiv 9^{52}\equiv 9^{102}\equiv \cdots$$

This means $$9 + \cdots + 9^{50} \equiv 9^{51}+\cdots + 9^{100} \equiv \cdots + 9^{1951} + \cdots + 9^{2000}\pmod{1000}$$

Noting $9^{51}\equiv 9\pmod{1000}$. $$9+9^2+\cdots + 9^{50} = \frac{9^{51}-9}{9-1}\equiv 0\times 8^{-1}\equiv 0\pmod{1000}$$

Therefore the desired answer is $40\times 0\equiv \boxed{0}\pmod{1000}$