[ฟินิกซ์เหินฟ้า]

ลองดูsoln(credit:AOPS)
Note here are $2012$ distinct nonzero integers of form $x-a_i$ if $x$ is an integer.
Also $(1006!)^2=1006.1005.......1.-1.-2......-1006$. Now if we pick one number among them and replace by a number greater than $1006$ or less than $-1006$ then certainly their product must be greater than $(1006!)^2$.
Thus we conclude we can write $(1006!)^2$ as product of $2012$ distinct nonzero integers in at most one way ,so done.