If $r$ is integer solution of $(x-a_1)(x-a_2)\cdots\cdots(x-a_{2n})+(-1)^{n-1}(n!)^2=0$ then $2nr=a_1+ a_2+\cdots+a_{2n}.$ Suppose that  $a_1, a_2,\ldots, a_{2n}$ are distinct integers such that 
$$(x-a_1)(x-a_2)(x-a_3)\cdots\cdots(x-a_{2n})+(-1)^{n-1}(n!)^2=0$$ has an integer root $r$. Show that
 $$2nr=a_1+ a_2+\cdots+a_{2n}.$$

You have  $|(r-a_1)| \cdot |(r-a_2) |\cdots |(r-a_{2n})|=(n!)^2$ with only strictly non negative integers.
Since all $a_i$ are distinct integers, so are all $r-a_i$.
The product of $n$ different strictly non negative integers is bounded by $1\cdot 2 \cdots n =(n!) $. Here this bound is reached two times so you have for all $1\le j \le n$ , exactly two different $a_i$ such that  $|r-a_i|=j$. And since all $r-a_i$ are distinct, the two numbers $r-a_i$ are the opposite numbers $j$ and $-j$.
In the end you have (by re ordering the terms) :
$$ (r-a_1) + (r-a_2) + \dots + (r-a_{2n}) = 1 -1 + 2-2 + \dots + n-n =0 $$ 
Which means $2nr = a_1 +a_2 + \cdots a_{2n}$.
