combinational proof from another side. How does using the group with the addition modulo I could solve this task:

Prove that among any $n + 2$ integers, there are two such that their
  sum or difference is divisible by $2n$.

 A: Modulo $2n$ the question can be rephrased as "In any set of $n+2$ residue classes (i.e. numbers), if they are all distinct (so that no difference is $0$), then there are two that add up to $0$."
It can be proven by pointing out that if you try to construct a counterexample, then for each number you include in your set, one other element is excluded from being picked. There just aren't enough residue classes to pick $n+2$ of them under that restriction.
Extended explanation
Well, what it basically is, is saying "I don't believe that it cannot be done. I'll make a set of size $n+2$ so that no sum or difference is divisible by $2n$." and then demonstrating, by keeping track modulo $2n$ that it is, indeed, impossible.
As I pick the first number into my counterexample set, say $a$, it's apparent that this excludes any number congruent to both $a$ and $2n-a$ from being picked at a later stage. Picking another number congruent to $a$ would make their difference divisible by $2n$, and picking a number congruent to $2n-a$ would make the sum divisible by $2n$.
Note that picking a number congruent to $0$ or $n$ only excludes one residue class from further picking. It's not important to the structure of the argument, but it's the reason the critical set size is $n+2$, not $n+1$.
As you keep on picking numbers, more and more residue classes modulo $2n$ will be completely unavailable to pick more numbers from. Finally, when you've picked $n+1$ numbers, all residue classes modulo $2n$ have been spent, and you cannot pick another number without forcing some sum or difference to be divisible by $2n$.
