Two wrongs do make a right

Consider any factorisation $100=(-a)(-b)$ , where $a \geq b$ are positive integers; $a$ can be 100, 50, 25, 20, or 10. If we add $S=a+b$ to each factors, the product will be $b\times a = 100$ once again. Now $S$ is smallest when $a=b=10$ , namely, $S=\boxed{20}$ .