Assassino9931 wrote: The finite set $M$ of real numbers is such that among any three of its elements there are two whose sum is in $M$. What is the maximum possible cardinality of $M$? Let $M^+=M\cap\mathbb R^+$ If $|M^+|\ge 4$, let $a>b>c>d>0$ the four greatest positive elements of $M$ Considering $\{a,b,c\}$, we have $a+b>a$ and so $\notin M$ and so $b+c>b\in M$ and so $c=a-b$ Considering $\{a,b,d\}$, we have $a+b>a$ and so $\notin M$ and so $b+d>b\in M$ and so $d=a-b=c$, impossible. So $|M^+|\le 3$ Let $M^-=M\cap\mathbb R^-$ If $|M^-|\ge 4$, let $a<b<c<d<0$ the four smallest negative elements of $M$ Considering $\{a,b,c\}$, we have $a+b<a$ and so $\notin M$ and so $b+c<b\in M$ and so $c=a-b$ Considering $\{a,b,d\}$, we have $a+b<a$ and so $\notin M$ and so $b+d<b\in M$ and so $d=a-b=c$, impossible. So $|M^-|\le 3$ So $\boxed{|M|\le 7}$, which indeed can be reached for example with $M=\{-3,-2,-1,0,1,2,3\}$