10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most?
10.5.7.

An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.

We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_{4}>0$ (since multiplying each number by -1 does not change our property). Thus, $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, so the sums $a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}$ do not belong to the set $A$. Furthermore, the sums $a_{2}+a_{3}$ and $a_{2}+a_{4}$ cannot both belong to the set $A$, because $a_{2}+a_{3}>a_{2}, a_{2}+a_{4}>a_{2}$, and $a_{2}+a_{3} \neq a_{2}+a_{4}$. Therefore, for the tuples $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(a_{1}, a_{2}, a_{4}\right)$, at least one of the tuples has the property that the sum of any two of its elements is not an element of $A$.