Number Theory: Prove $a_1 = a_2 = ... = a_{mn+1}$ 
We have a set of integers $a_1, a_2, ..., a_{mn+1}$. If we put one
  number aside, we can divide the rest to $m$ groups each with $n$ number in a
  way that sum of numbers in each group is same. Prove that $a_1 = a_2 =$
  $... = a_{mn+1}$

How can i prove this statement?
 A: Let $i$ be such that $a_i$ is the smallest number in the set. Let us now consider the set $\{a_1-a_i,\dots,a_{mn+1}-a_i\}$: every number in this set is non-negative, and at least one number is $0$, and this set clearly has the same property the original set had. Let us put $b_j:=a_j-a_i$, so that $b_i=0$. 
Let $S:=\sum_{j=1}^{mn+1}b_j$. If we remove $b_i=0$, by the property of the set we deduce that $S$ is divisible by $n$. If we remove any other $b_j$, then by the property $S-b_j$ is divisible by $n$, which implies that every $b_j$ is divisible by $n$. We can then consider the set $\{b_1/n,\dots,b_{nm+1}/n\}$, which again has the property of the initial set. Since we can repeat this argument an infinite number of times, the $b_j$ had to be all equal to $0$.
A: I am going to build on @Leo163's answer. 
Assume no two $a_i$ are the same. 
Consider the set $(b_1, b_2, ..., b_i, ... b_{mn+1})$ and assume WLOG $b_i = 0$. 
Let $N = \sum_{i=1}^{mn+1}b_i$. 
Let $S_k$ represent the sum of the numbers in the groups when $b_k$ is removed. 
$N = mS_1 + b_1$ 
$N = mS_2 + b_2$ etc except 
$N = mS_i$ as $b_i = 0$
So we can see that $m \mid b_k$ for $1 \le k \le mn+1$
Now we can construct another set $(b_1/m, b_2/m,...,b_i/m, ... b_{mn+1}/m)$ with similar properties such as the original set. 
We can keep going on this infinite descent $\implies b_k = 0$ for $1 \le k \le mn+1$
$\implies a_i = a_j$ for $i \ne j$ 
