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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.

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    Where is this question from? Also, what have you tried? For instance, it could be fruitful to try out with $7$ numbers so that any $6$ of them can be partitioned into two sets of $3$ such that the sum is equal. It's a lot easier to keep track of everything, and you might spot a pattern that generalises. On the other hand, it could be that this statement doesn't _hold_ for such a small set... – 2017-01-28
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    Where did you find this problem? Combinatorics and number-theory problems that involve the current year are often contest problems, so many readers will refrain from giving too complete replies, lest they become complicit in cheating. – 2017-01-28
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    Well the number 2017 doesn't have any particular meaning. It just has to be an odd number and is quite trivial for $n=3$ – 2017-01-28
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    @HenningMakholm I have encountered a similar problem with $n=23$ some time ago, so I doubt this is problem from a current contest. Unfortunately I can't recall the solution at the moment. – 2017-01-28
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    No, it's not from any current contest. I just put that number there. It might have been given on a contest before. – 2017-01-28

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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 $0$.