Prove that the elements of X all have the same weight I have had this problem on my mind for weeks and haven't been able to find a solution. Let $X$ be a set with $2n+1$ elements, each of which has a positive "weight" (formally, we could say there exists a weight function $w:X\to \mathbb{R}^+$). Suppose that for every $x\in X$, there exists a partition of $X\setminus\{x\}$ into two subsets each containing $n$ elements such that the two sums of the weights of the elements in each subset are equal (or using mathematical symbols, $\forall x\in X$, $\exists Y,Z\subset X$, such that $Y\cup Z=X\setminus\{x\}$, $|Y|=|Z|=n$, and $\sum_{y\in Y}w(y)=\sum_{z\in Z}w(z)$). Prove that the elements of $X$ must have the same weight.
The converse is obvious. If the elements of $X$ have the same weight, then clearly any partition of $X\setminus\{x\}$ into two subsets of equal size will suffice. It's this direction that's troublesome. I have tried using induction, and I have tried turning it into a linear algebra problem, but I haven't had luck with either of these methods. If anyone can think of a solution, I'd love to hear it.

The weights $w_i$ span a vector space $V:=\langle w_1,w_2,\ldots,w_{2n+1}\rangle$ over ${\mathbb Q}$ of dimension $d\leq2n+1$. Let $(\xi_1,\ldots,\xi_d)$ with $\xi_k\in{\mathbb R}$ be a basis of $V$. Then there are rational numbers $a_{ik}$ such that
$$w_i=\sum_{k=1}^d a_{ik}\,\xi_k\qquad(1\leq i\leq 2n+1)\ .$$
Any integer relation $$\sum_{i=1}^{2n+1} n_i\,w_i=0\tag{1}$$
among the $w_i$ then implies $\sum_{k=1}^d\left(\sum_{i=1}^{2n+1}n_i a_{ik}\right)\xi_k=0$, hence
$$\sum_{i=1}^{2n+1}n_i a_{ik}=0\qquad(1\leq k\leq n)\ ,$$
saying that each $(2n+1)$-tuple ${\bf a}_k:=(a_{ik})_{1\leq i\leq 2n+1}$ satisfies $(1)$. This allows to conclude that each ${\bf a}_k$ $(1\leq k\leq d)$ satisfies the premises of the question. After dividing $\xi_k$ by the LCM of the $a_{ik}$ we may assume that all $a_{ik}$ are integers.
We now make use of  @joriki 's elegant argument to show that the ${\bf a}_k$ are in fact constant. So there are numbers $m_k$ $(1\leq k\leq d)$ with $a_{ik}=m_k$ $(1\leq i\leq 2n+1)$. It follows that
$$w_i=\sum_{k=1}^d m_k\,\xi_k\qquad(1\leq i\leq 2n+1)\ .$$
