jmc

If $x^2+x-n$ factors as the product of two linear factors with integer coefficients, then it must be of the form $$(x-a)(x-b)=x^2-(a+b)x+ab,$$where $a$ and $b$ are integers. Then $a+b=-1$ and $ab=-n$, which means $n=-ab=-a(-a-1)=a(a+1).$ We want $1\le n\le 1000.$ The possible values of $a$ are then 1, 2, $\dots ,$ 31, so there are $\boxed{31}$ possible values of $n.$ (Note that $a$ can also be $-32,$ $-31,$ $\dots ,$ $-2,$ but these give the same values of $n.$)