It is said that the following proposition is true.


$\forall S \subset \mathbb{Z}, |S| = 2n-1.\ \exists A \subset S, |A| = n$ which satisfies 
$$
n \ | \ \sum\_{a \in A}a.
$$
Could someone gives a proof or some infomation about this? Thx.



Find a subset such that its sum is divisible by $n$
It is [Erdős–Ginzburg–Ziv theorem](https://en.wikipedia.org/wiki/Zero-sum_problem). 
The standard proof is induction on number of prime divisors of $n$. 


If $n$ is prime, let $0\leqslant x\_1\leqslant \dots \leqslant x\_{2n-1}\leqslant n-1$ be elements of $S$ (we may think so of course). If $x\_i=x\_{n-1+i}$ for some index $i\in \{1,2,\dots,n-1\}$, we take $A=\{x\_i,\dots,x\_{n-1+i}\}$. If not, denote sets $U\_i:=\{x\_i,x\_{n-1+i}\}$ for $i=1,\dots,n-1$, $U\_0:=\{x\_{2n-1}\}$. I claim that $|U\_0+\dots+U\_k|\geqslant k+1$ for all $k=0,\dots, n-1$ (set addition is modulo $n$), for $k=n-1$ this is implies the claim, since $0\in U\_0+\dots+U\_{n-1}$ is enough for our goal. This is easy induction. For $k=0$ the claim is clear. If it holds for $k-1$, but does not hold for $k$, then the sets $A=U\_0+\dots+U\_{k-1}+x\_k$ and $B=U\_0+\dots+U\_{k-1}+x\_{n-1+k}$ each have exactly $k$ elements and coincide. But $B=A+(x\_{n-1+k}-x\_k)$, and, since $n$ is prime, there is no subset $A$ of residues mod $n$, which has $k$ elements and coincides with its translate $A+t$, $t\ne 0$. 


(Remark: this step is often proved using [Cauchy-Davenport theorem](https://en.wikipedia.org/wiki/Restricted_sumset), but you see that we need only an easy particular case of Cauchy-Davenport.)


Now induction step. If $n=ab$, $a>1$, $b>1$ and both for $a$ and for $b$ the claim holds, then we may find $a$ elements with sum divisible by $a$ out of any $2a-1$ elements. Do this, remove $a$ these elements and proceed. We may do this $2b-1$ times, i.e. there exist $2b-1$ disjoint subsets $A\_1,\dots,A\_{2b-1}$, $|A\_i|=a$, $\sum\_{a\in A\_i}a=ak\_i$. Apply the claim for $k\_1,\dots,k\_{2b-1}$.


Remark: almost the same argument works in any Abelian group of order $n$. 


