jmc

We claim that the minimum value is $-101.$

If $a=-1$ and $b=-100,$ then $ab=100$ and $a+b=-101.$

Now, $$\begin{array}{rl}a+b+101& =a+\frac{100}{a}+101\\ & =\frac{a^2+101a+100}{a}\\ & =\frac{(a+1)(a+100)}{a}.\end{array}$$If $a$ is positive, then $b$ is positive, so $a+b$ is positive, so suppose $a$ is negative. Then $b$ is negative. Furthermore, since $a$ is a factor of 100, $-100\le a\le -1.$ Hence, $a+1\le 0$ and $a+100\ge 0,$ so $$a+b+101=\frac{(a+1)(a+100)}{a}\ge 0.$$Equality occurs if and only if $a=-1$ or $a=-100,$ both of which lead to $a+b=-101.$

Therefore, the minimum value of $a+b$ is $\boxed{-101}.$