Use Pigeonhole to show of any set of $2^{n+1}-1$ positive integers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$. 
Show that, of any set of $2^{n+1}-1$ positive integers numbers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$.

My approach: Let $a_1,a_2,\ldots,a_{2^{n+1}-1}$ positive integers numbers. Define, $s_j=a_1+a_2+a_3+\ldots+a_{2^{j+1}-1}$, $j\in\{1,\ldots,n\}$. 
If any of the $2^{n+1}-1$, $s_j$, are divisible by $2^n$, we are done. 
In case contrary, $s_j$ have a remainder $1,2,3,\ldots,2^{n}-1\mod(2^n)$. And, we know that $2^{n+1}-1>2^n-1$, so by Pigeonhole Principle, there exist a tleast two $s_j$ such that has a same remainder ($\mod(2^n)$), i.e., if $p<q$, then $$s_q-s_p=a_{2^{p+1}}+a_{2^{p+1}+1}+\ldots+a_{2^{q+1}-2}+a_{2^{q+1}-1}\equiv 0\mod(2^{n})$$
So, the set $\{a_{2^{p+1}},a_{2^{p+1}+1},\ldots,a_{2^{q+1}-2},a_{2^{q+1}-1}\}$ is the set we need. I know, this answer doesn't be right, but is the idea I have to occupy Pigeonhole.
EDIT: Both answers were very good, but, how I could occupy the pigeonhole principle in this problem?

As suggested in the comments this is Erdos-Ginzburg-Ziv theorem for a power of $2$.
The proof of E-G-Z has an easy inductive part, that if the theorem is true for $p$ and $q$ it is true for $pq$, and a nontrivial base case of proving the theorem true for primes.  Here the prime is $2$ so the base case is true by inspection. The proof of the  multiplicative property is the same whether or not $p=q$ or if they are powers of $2$.
https://en.wikipedia.org/wiki/Zero-sum_problem
