Let the roots be $a$ and $b$. We have $a+b=-\frac{r+1}r\in\mathbb Z$ and $ab=\frac{r-1}r$. We need $\frac1r$ to be an integer. Let $r=\frac1n$. It becomes $x^2+(1+n)x+1-n=0$, so $a+b+1=-n$ and $ab-1=-n$, so $a+b+1=ab-1$. It factors as $3=(a-1)(b-1)$, so $a\in\{-2,0,2,4\}$. At this point, we just bash divisors.

Also note that $r=0$ works. I think this type of value is missed a lot in these problems because competitors automatically assume it must be a quadratic.