jmc

We use the fact that the sum and product of the roots of a quadratic equation $a{x}^2+bx+c=0$ are given by $-b/a$ and $c/a$, respectively. Let the two roots of the equation be $p$ and $q$. Then $p+q=k/2$. However the only other restriction on $p$ and $q$ is that $pq=4$ and that $p$ and $q$ are distinct integers. For each such possibility $(p,q)$, we also have the possibility $(-p,-q)$ since $(-p)(-q)=pq=4$. This gives two values of $k$: $k=2(p+q)$ and $k=2(-p-q)$. Since these occur in such pairs, the sum of all possible values of $k$ is $\boxed{0}$.

Alternatively, one can note that the only way to factor 4 into 2 distinct integer factors is $4\cdot 1$ and $(-4)(-1)$, so that the two possible values of $k$ are $10$ and $-10$, given a sum of $0$.