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.
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 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 :
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.