Remove any number and the remaining numbers can be partitioned into two subsets of equal sum; prove all numbers are equal.

Question

Supposed I have a list of $n$ real numbers, where $n$ is odd. The list is constructed such that I can remove any arbitrary number from the list, and the remaining numbers can be partitioned into two equal-sized subsets with equal sums. Prove that all numbers in the list are equal.

This should be somehow related to linear algebra. A way I could think of to interpret this is that the list is essentially a $1 \times n$ row, and there exist $n$ $n \times 1$ vectors with one zero in some entry and $1$'s and $-1$'s in other entries with the entries summing to $0$, such that the product of the row and the column vector is $[0]$.

In other words, the entries/$1 \times 1$ columns in the row are linearly dependent once we remove any arbitrary entry/column. I'm not sure how this is/could be related to the proof though.

Thanks in advance!

Answer 1

Let $x=(x_1,x_2,\dots,x_n)$ be the vector of real numbers.

For each $1\le i \le n$, there exist disjoint sets $P_i,Q_i\subset \{1,2,\dots,n\}$ whose union is $\{1,2,\dots,n\}\setminus \{i\}$ such that $\sum_{j\in P_i}x_j=\sum_{j\in Q_i}x_j$. Let $M$ be the $n\times n$ matrix defined by
$$ M_{i,j}=\begin{cases} 1 & x_j\in P_i\\ -1 & x_j\in Q_i\\ 0 & i=j\end{cases} $$ Letting $\bf 1$ be the vector of all ones, then we have that $$ Mx=0\qquad \text{and}\qquad M{\bf 1}=0 $$ If we can show that the null space of $M$ has dimension $1$, this will prove $x$ is a scalar multiple of $\bf 1$, so all weights are equal.

It suffices to prove that $M'$, the upper $(n-1)\times (n-1)$ submatrix of $M$, is invertible. Letting $J$ be the $(n-1)\times (n-1)$ matrix of all ones, and $I$ be the $(n-1)\times (n-1)$ identity, then each element of $M'$ is congruent mod $2$ to the corresponding element of $J-I$, so $\det M'\equiv \det(J-I)\pmod 2$. Considering $J-I$ as a matrix over $\mathbb F_2$, we have (noting each entry of $J^2$ is $n-1\equiv 0\pmod 2$) $$ (J-I)^2 = J^2-2JI+I^2=I, $$ which proves $J-I$ is invertible mod $2$, so $\det(J-I)\equiv 1\pmod 2$. This proves $\det M'\equiv 1\pmod 2$, so $\det M'$ is nonzero, as required.