1,000

$\sum_{n=1 }^{ 1000} n =\frac{n(n+1)}{2} =\frac{1000(1001)}{2} =500500$

This is a sum of arithmetic progression , which is given by $\displaystyle S(n) = \sum_{k=1}^n k = \frac {n(n+1)}2$ . For $n=1000$ , $S(1000) = \dfrac {1000\times 1001}2 = \boxed{500500}$ .

$\frac{1000+1}{2} \times 1000$