Start Counting! Part-3

$x+1$ divides $ax^2+bx+c$ when $a(-1)^2+b(-1)+c=a-b+c=0$ . Given $a\in\{1,2,\ldots,1999\}$ , there are $1999-a$ values of $c$ ( $1,2,\ldots,1999-a$ that yield $1\le b\le 1999$ . If $a\le 1999-a$ (i.e. $a<1000$ ) then we must further rule out $a$ as a value for $c$ .

Therefore, the answer is

$\displaystyle\sum_{a=1}^{999}(1998-a)+\sum_{a=1000}^{1999}(1999-a)=999\cdot1998+1000\cdot 1999-\sum_{a=1}^{1999}a$

$\displaystyle=999\cdot1998+1000\cdot 1999-\frac{1}{2}1999\cdot2000=999\cdot1998=\boxed{1996002}$ .