Answer: 600

Solution:

The polynomial factors into integer linear factors exactly when its roots are integers, that is, when the discriminant $a^2 - 4b$ equals $c^2$ for some integer $c \ge 0.$ Given $a,$ such a $b \ge 0$ exists exactly when $4b = a^2 - c^2 = (a-c)(a+c)$ for some $c$ with $0 \le c \le a$ and $c \equiv a \pmod 2,$ and distinct such $c$ give distinct values $b = \frac{a^2 - c^2}{4}.$

For odd $a$ the valid $c$ are $1, 3, \ldots, a,$ which is $\frac{a+1}{2}$ choices; for even $a$ they are $0, 2, \ldots, a,$ which is $\frac{a}{2} + 1$ choices.

Summing over $a = 1, \ldots, 100:$ the odd $a$ contribute $1 + 2 + \cdots + 50 = 1275,$ and the even $a$ contribute $2 + 3 + \cdots + 51 = 1325.$ Thus $S = 2600,$ and the remainder is $600.$