Playing with a simple rational function!

First, you need to notice that we can assume that $|a|\neq|b|,$ because if $|a|=|b|,$ then $f(x)=\pm \frac{ b}{x-1}$ or $f(x)=\pm\frac{ b}{x+1},$ and then, using that $b\neq 0,$ there cannot be two different numbers that do not belong to the range of $f(x).$ Using the previous assumption, we obtain that the equation $y=\frac{ax+b}{x^2-1}$ is equivalent to $y(x^2-1)=(ax+b),$ because $x=1$ and $x=-1$ would not be solutions of the latter equation. Therefore,

$y\in Range(f)$

$\Leftrightarrow y(x^2-1)=(ax+b)\quad\text{has real solutions with respect to}\quad x \quad\text{(distinct from 1 and -1)}$

$\Leftrightarrow yx^2-ax-(y+b)=0\quad \text{has real solutions with respect to}\quad x \quad \text{(distinct from 1 and -1)}$

$\Leftrightarrow a^2 +4y(y+b)\geq 0 \quad \text{(discriminant)}$

$\Leftrightarrow (2y+b)^2\geq b^2-a^2$

From the previous equivalences, if $|b|<|a|,$ then $Range(f)= (-\infty, \infty).$ Otherwise, in the case that $|b|>|a|,$ we have that $Range(f)= (-\infty, \frac{-b-\sqrt{b^2-a^2}}{2}]\cup [ \frac{-b+\sqrt{b^2-a^2}}{2}, \infty).$ Using the fact that 1 and 100 are not in the range of $f$ , we obtain that $\frac{-b-\sqrt{b^2-a^2}}{2}<1, \text{ and} \ \frac{-b+\sqrt{b^2-a^2}}{2}>100 \quad \quad\quad (*)$ The second of these two inequalities and the fact that $\sqrt{b^2-a^2}\leq |b|,$ imply that $\frac{-b+|b|}{2} \geq\frac{-b+\sqrt{b^2-a^2}}{2}>100,$ so $b$ is a negative number and $|b|>100,$ and, since $b$ is integer, $|b|\geq 101.$ Then, $b=-101$ . Using this value of $b$ , we can find the possible values of $a$ solving either inequality in (*) for $|a|$ , In both cases, we obtain that $|a|<20$ . Therefore, the possible pairs for which $b$ takes its minimum absolute value are $( -19, -101), (-18, -101), ..., (19, -101),$ and the answer for this question would be $\boxed{39}$ .