9.5. Let $M$ - be a finite set of numbers (distinct). It is known that among any three of its elements, there will be two whose sum belongs to $M$. What is the maximum number of elements that can be in $M$?
Answer: 7.

Solution: An example of a set with 7 elements: $\{-3,-2,-1,0,1,2,3\}$.

We will prove that a set $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $n>7$ 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 $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, i.e., 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 $a_{2}+a_{3}>a_{2}+a_{4}>a_{2}$. This means that for at least one of the triples ( $a_{1}, a_{2}, a_{3}$ ) and ( $a_{1}, a_{2}, a_{4}$ ), the sum of any two of its elements does not belong to the set $M$.

Grading criteria: Proved that there are no more than 7 elements in M: 5 points. Example for 7 elements: 2 points.