Let $M$ be a finite set of numbers. It is known that among any three of its elements, there are two whose sum belongs to $M$.

What is the maximum number of elements that can be in $M$?
Consider either the four largest or the four smallest numbers.

## Solution

Example of a set of 7 elements: $\{-3,-2,-1,0,1,2,3\}$.

We will prove that the set $M=\left\{a_{1}, a_{1}, \ldots, a_{n}\right\}$ of $n>7$ numbers does not have the required property. We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a_{4}>0$ (changing the signs of all elements does not change our property). Then $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, which means that none of the sums $a_{1}+a_{2}, a_{1}+a_{3}$, and $a_{1}+a_{4}$ belong to the set $M$. Moreover, the sums $a_{2}+a_{3}$ and $a_{2}+a_{4}$ cannot both belong to $M$, since $a_{2}+a_{3}>a_{2}+a_{4}>a_{2}$. Therefore, for at least one of the triples $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(a_{1}, a_{2}, a_{4}\right)$, the sum of any two of its elements does not belong to the set $M$.

## Answer

7 elements.