解:①$2 + 4 + 6 + \cdots + 2000,$由于$2000 = 2×1000,$根据规律可得和为$1000×(1000 + 1) = 1000×1001 = 1001000。$②$202 + 204 + \cdots + 600,$先求$2 + 4 + \cdots + 600$的和($n = 300$):$300×301 = 90300;$再求$2 + 4 + \cdots + 200$的和($n = 100$):$100×101 = 10100;$两者相减得$90300 - 10100 = 80200。$