Polynomial olympiad problem

Observe that $\{r-a, r-b, r-c, r-d \}=\{\pm 1, \pm 2\}$. Thus $$r-a + r-b + r-c + r-d=4r-(a+b+c+d)=0$$

As you said, we have $4=(r-a)(r-b)(r-c)(r-d)$. Therefore, as $a,b,c,d$ are integers. the absolute values of any two of the factors of $4$ above should be $2$ and two of them should be $1$. Now, the numbers which have the same absolute values must have different sign, as otherwise, the numbers would become equal. Thus, let us assume, without loss of generality $r-a=2, r-b=-2,r-c=1,r-d=-1$, which gives us $a+b+c+d=4r$, from which the desired conclusion follows.

Since a,b,c and d are distinct, we can suppose that $a<b<c<d$. The only possibility to have $$4=(r−a)(r−b)(r−c)(r−d) $$ is when $a<b<r<c<d$ and when $$r-a=2,r-b=1,r-c=-1,r-d=-2.$$ In this configuration, one has $r-a+r-b+r-c+r-d=4r-(a+b+c+d)=0$ and then $$r=\frac{a+b+c+d}{4}.$$