jmc

Since $x$ and $1/x$ are integers, we know that $x$ is an integer that divides 1 evenly. Therefore, we must have $x=-1$ or $x=1$. The same is true for $y$ and $z$. So, the possible sums are $3(-1)=-3$, $2(-1)+1=-1$, $2(1)+-1=1$ or $3(1)=3$. So there are $\boxed{4}$ possible values for the sum.