74th Romanian Mathematical Olympiad

a) Since $1+6=2+5=3+4$, every set of $5$ numbers from $1,2,3,4,5,6$ contains $4$ distinct numbers $a,b,c,d$ so that $a+b=c+d=7$.

b) Indeed, no $4$ numbers out of $1,2,3,5,8$ provide equal sums: if we do not choose $8$, then $5+a>b+c$, for every $a,b,c\in \{1,2,3\}$, and if we choose $8$, then $8+a>b+c$, for every $a,b,c\in \{1,2,3,5\}$. The number $n=7$ is special. Indeed: if we do not choose $7$, $4$, or $1$, then the argument from a) applies to the sums $1+6=2+5=3+4$, $1+7=2+6=3+5$, respectively $2+7=3+6=4+5$; if we choose $1$, $4$, or $7$ and two of the numbers $2,3,5,6$, then we get the equal sums $1+4=3+2$, $1+5=2+4$, $1+7=2+6$, $1+7=3+5$, $3+7=4+6$, or $4+7=5+6$. Since $5$ is, obviously, special, the special numbers are $5$, $6$ and $7$.