$12|(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$ (this was posted many times, even recently so I won't be proving this, anyway it's straightforward), so we're need to find $a,b,c,d$ such that : $(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)=12$ take: $a=b+1$, $b-c=2$, $b-d=1$ and it is easy to check that in this case indeed $(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)=12$ so we're done.

Do You have any idea how to kill this problem in a quick way?

Yes. For every integer $ n$ greater than $ 4$ we may find a set of numbers $ \{a,b,c,d\}$ such that every one of them has different residue $ mod \ n$ The conclusion follows...