jmc

By the quadratic equation, the roots of the equation are $$\begin{array}{rl}\frac{-b\pm \sqrt{b^2-4ac}}{2a}& =\frac{-s\pm \sqrt{s^2-4(\frac{1}{2})(-\frac{1}{2})}}{2(\frac{1}{2})}\\ & =\frac{-s\pm \sqrt{s^2+1}}{1}=-s\pm \sqrt{s^2+1}.\end{array}$$ Thus we know that $-s+\sqrt{s^2+1}$ and $-s-\sqrt{s^2+1}$ are integers. We know that $s$ is an integer, so in order for the sum to be an integer, we must have that $\sqrt{s^2+1}$ is an integer.

Let $\sqrt{s^2+1}=n$ for some integer $n$. Then we have $s^2+1=n^2$, or $n^2-s^2=1$ and thus $$(n-s)(n+s)=1.$$ Since $n$ and $s$ are both integers, then their sum and difference must also both be integers, so they are either both $1$ or both $-1$ since their product is 1. In either case, $n-s=n+s$, so $2s=0$ and $s=0$. This is the only value of $s$ for which $\sqrt{s^2+1}$ is an integer, and thus the only value of $s$ that makes the roots of the given quadratic integers, so $s=\boxed{0}$.