Proving that cows have same weight without weighing it! My friend gave me this problem and I have no clue how to go about it:

A peasant  has $2n + 1$ cows. When he puts aside any of his cows,
the remaining $2n$, can be divided into two sub-flocks of $n$ cows and each having
the same total weight. How can we prove that the cows all have the same weight? 

How can I approach this?
Let $w_i$ denote the weight of cow $i$, $1\le i\le 2n+1$. 
With $\mathbf e_i$ the $i$th member of the standard basis of $\mathbb R^{2n+1}$
let  $\mathbf w=\sum_i w_i\mathbf e_i$ and $\mathbf j=\sum_i \mathbf e_i$.
By assumption, for each $k$, $1\le k\le 2n+1$, there exists a vector $\mathbf a_k=\sum_ia_{k,i}\mathbf e_i\in \mathbb Z^{2n+1}\subset\mathbb R^{2n+1}$ with 


$a_{k,k}=0$, 
$a_{k,i}\in\{-1,1\}$ for all $i\ne k$, 
$\langle \mathbf a_k,\mathbf j\rangle=0$,
$\langle \mathbf a_k,\mathbf w\rangle = 0$. 


Then the vectors $\mathbf a_k+\mathbf j\equiv \mathbf e_k \pmod 2$, i.e., the $\mathbf a_k+\mathbf j+2\mathbb Z^{2n+1}$ are the standard basis of $\mathbb Z^{2n+1}/2\mathbb Z^{2n+1}\cong \mathbb F_2^{2n+1}$. 
Assume $\sum_k q_k(\mathbf a_k+\mathbf j)=0$ with $q_k\in\mathbb Q$, not all $=0$. By multipying with the common denominator we may assume that all $q_k\in\mathbb Z$, by dividing out common factors we may assume at least one of the $q_k$ is odd. But then taking remainders $\bmod 2$ produces a linear dependence among the $\mathbf a_k+\mathbf j+2\mathbb Z^{2n+1}$. From this contradiction, we conclude that  the  $\mathbf a_k+\mathbf j$ are $\mathbb Q$-linearly independent and hence a basis of $\mathbb Q^{2n+1}\subset\mathbb R^{2n+1}$. 
Thus we can write $\mathbf e_k=\sum_i c_{k,i}(\mathbf a_i+\mathbf j)$ with $c_{k,i}\in \mathbb Q$ and have
$$ 1=\langle \mathbf e_k,\mathbf j\rangle=\sum_ic_{k,i}(\langle \mathbf a_i,\mathbf j\rangle +\langle\mathbf j,\mathbf j\rangle)=(2n+1)\sum_ic_{k,i}$$
and hence 
$$ \begin{align}w_k&amp;=\langle \mathbf e_k,\mathbf w\rangle\\&amp;=\sum_i c_{k,i}\langle \mathbf a_i,\mathbf w\rangle +\sum_i c_{k,i}\langle \mathbf j,\mathbf w\rangle\\
&amp;=\rlap{\qquad0}\hphantom{\sum_i c_{k,i}\langle \mathbf a_i,\mathbf w\rangle}+\langle \mathbf j,\mathbf w\rangle\cdot \sum_ic_{k,i}\\&amp;=\frac{\langle \mathbf j,\mathbf w\rangle}{2n+1}\end{align}$$
which does not depend on $k$.
We have $2n+1$ equations in $2n+1$ unknowns. This has a unique solution. As all the cows having the same weight is a solution, it is the only solution.