Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$.
Let the linear factors be $(x+c)(x+d)$.
Then, $a=c+d$ and $b=cd$.
We know that $1\le a\le 100$ and $b\ge 0$, so $c$ and $d$ both have to be non-negative
However, $a$ cannot be $0$, so at least one of $c$ and $d$ must be greater than $0$, ie positive.
Also, $a$ cannot be greater than $100$, so $c+d$ must be less than or equal to $100$.
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices $(0,0), (0, 100),$ and $(100,0)$. Remember that $(0,0)$ does not work, so there is a square with the top right corner $(1,1)$.
Note that $c$ and $d$ are interchangeable since they end up as $a$ and $b$ in the end anyways. Thus, we simply draw a line from $(1,1)$ to $(50,50)$, designating one of the halves as our solution (since the other side is simply the coordinates flipped).
We note that the pattern from $(1,1)$ to $(50,50)$ is $2+3+4+\dots+51$ solutions and from $(51, 49)$ to $(100,0)$ is $50+49+48+\dots+1$ solutions, since we can decrease the $y$-value by $1$ until $0$ for each coordinate.
Adding up gives \[\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.\]
This gives us $2600$, and $2600\equiv 600 \bmod{1000}.$
Thus, the answer is: \[\boxed{600}.\]
~Minor edit by Yiyj1