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.
Best Answer
Assume the elements are not all the same. Multiply them all by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $\frac1{2n}$ from the nearest integer. These rescalings preserve the premise. Since the differences from the nearest integers add up to less than $1$, the condition in the premise must hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integer elements.
So assume integer elements. Subtract the smallest element from all elements; this preserves the premise. Now one element is $0$. If they're not all $0$, divide through by the greatest power of $2$ that divides all of them; this also preserves the premise. Now one of them is odd and the $0$ is even; but the premise implies that they're either all even or all odd; a contradiction; hence they were all $0$ and thus originally all equal.