Let $r, s$ be the roots of $x^{2}-a x+a=0$. By Vieta's, we have $r+s=r s=a$. Note that if one root is an integer, then both roots must be integers, as they sum to an integer $a$. Then, $$r s-(r+s)+1=1 \Longrightarrow(r-1)(s-1)=1$$ Because we require $r, s$ to be both integers, we have $r-1=s-1= \pm 1$, which yields $r=s=0,2$. If $r=0$ or $s=0$, then $a=0$, but we want $a$ to be a positive integer. Therefore, our only possibility is when $r=s=2$, which yields $a=4$, so there is exactly 1 value of $a$ (namely, $a=4$ ) such that $x^{2}-a x-a$ has an integer root.