62nd Ukrainian National Mathematical Olympiad

Answer: $n=2$.

Suppose that there are at least three numbers in the set $M$, denote them by $a<b<c$. By the condition $b^2+1⊢c$, and therefore the numbers $b,c$ are mutually prime. Since $a^2+1⊢b$ and $a^2+1⊢c\Rightarrow$ $$a^2+1⊢bc\ge (a+1)(a+2)=a^2+3a+2.$$ We obtained a contradiction that proves that $n\le 2$, and for $n=2$ it is enough to consider, for example, the set $M=\{1,2\}$.