Equality of elements of set Suppose you have a set of 2n+1 real numbers with the property that taking any one element out of the set, you can arrange the remaining 2n elements into two groups (each group having n elements only)  of equal sum. Prove all elements are equal.
I tried proving it using induction. 
Suppose it is true for some $2n+1$.
Then I tried to prove that it must be true for $2(n-1)+1$ elements also. The condition is easily established for 3 elements.  So I thought that induction could work.  But I was unable to do so. 
Any hint regarding how to prove this using induction or another method would be appreciated. 
 A: This solution was communicated to me by Bruno Grebille, from Toulouse.
Let $X=\begin{pmatrix}x_1\\\vdots\\x_{2n+1}\end{pmatrix}$ be your $2n+1$ reals.
By hypothesis, there exists a matrix $A=(a_{i,j})\in M_{2n+1}$ such that :

*

*the diagonal elements $a_{i,i}$ are $0$ ;

*apart from them, each line consists of exactly $n$ elements equal to $1$ and $n$ elements equal to $-1$ ;

*$AX=0$.

The kernel of such a matrix clearly contains the vector $U=\begin{pmatrix}1\\\vdots\\1\end{pmatrix}$ hence in order to prove that $X$ is collinear to $U$, we just have to show that $\operatorname{rank}(A)$ equals $2n$ (and not less).
Since $A\in M_{2n+1}(\mathbb Q)$, we are reduced to proving the initial claim with the extra hypothesis that all $x_i$'s are integers. Let us do this by contradiction, assuming that the differences $x_i-x_j$ are not all $0$, and choosing such a difference having minimal $2$-valuation :

*

*$\forall i,j\quad2^p\mid x_i-x_j$,

*$2^{p+1}\not\mid x_1-x_2$ (wlog).

Wlog (changing the first line of $A$ to its opposite if necessary) $a_{1,2}=a_{2,1}$. Then, the difference of the first two equations in $AX=0$ gives
$a_{1,2}(x_1-x_2)=\sum_{j\ge3}b_jx_j$ with each $b_j=a_{1,j}-a_{2,j}$ equal to $0$, $2$, or $-2$, and $\sum_{j\ge3}b_j=0$, hence
$$2^{p+1}\mid b_3(x_3-x_4)+(b_3+b_4)(x_4-x_5)+(b_3+b_4+b_5)(x_5-x_6)+\dots+(b_3+\dots+b_{2n})(x_{2n}-x_{2n+1})=\sum_{j\ge3}b_jx_j=\pm(x_1-x_2).$$
This contradiction ends the proof.
