Partitions of 2017 natural numbers

Question

Suppose we have 2017 natural numbers, such that each 2016 can be grouped into 2 groups with equal sum and equal number of elements. Prove that all numbers are equal.

Answer 1

Let us call $S$ the set of numbers we have.Let $a$ be the minimum of $S$. If we subtract $a$ to any number we have, the property still holds. This means that we can consider $S$ containing $0$.

We claim that every number in $S$ is even. Suppose not, and let $b\in S$ be odd. The sum of all number in $S$ is even, since it is $0+2c$ for some $c$ (by the hypothesis on $S$). Then, we should have $2c=b+2d$ for some $d$, which is impossible. Then, all elements of $S$ are even.

If we divide every element of $S$ by $2$, the new set $S'$ we obtain still has the property and contains $0$. This means that we can iterate the argument above. It can only be done if all elements of $S$ are $0$.