The polynomial of this kind is factorable IF and ONLY IF  " n " is the product of two consecutive integer numbers n = m*(m+1).


Then  x^2 + x -n = (x+m)*(x-(m+1)).


So, you should determine the number of all such positive n's  under the condition  1 <= m*(m+1) <= 1000.


Notice that " m " can be positive and negative, as well.


So, you need to find the number of solutions to this compound inequality

        1 <= m*(m+1) <= 1000.


The solutions are   -32 <= m <= 31; but you should reject negative m's and m = 0.

So, the number of such positive n's is 31.