jmc

Because the given quadratic has leading coefficient $1$, both factors must be of the form $x-c$ (or $-x+c$). Therefore, such a factorization exists if and only if $x^2+ax+b$ has two integer roots. Letting $r$ and $s$ denote these roots, we have, by Vieta's formulas, $$\begin{array}{rl}r+s& =-a,\\ rs& =b.\end{array}$$Since $r+s=-a$ is negative but $rs=b$ is nonnegative, it follows that both $r$ and $s$ must be negative or zero. Now, for each $a$, there are $a+1$ possible pairs $(r,s)$, which are $(0,-a)$, $(-1,-a+1)$, $\dots$, $(-a,0)$. However, since the order of $r$ and $s$ does not matter, we only get $\lceil \frac{a+1}{2}\rceil$ distinct polynomials $x^2+ax+b$ for each possible value of $a$. It follows that the number of these polynomials is $$\sum _{a=1}^{100}\lceil \frac{a+1}{2}\rceil =1+2+2+3+3+\cdots +50+50+51=\boxed{2600}$$since if we pair up the terms in this sum end-to-end, each pair has a sum of $52=2\cdot 26$.