Romanian Mathematical Olympiad

Observe that $A_n$ has $n^2+n+1$ elements, hence contains all the elements of $M$ except one. Similarly, $A_{n-1}$ contains all the elements of $M$ except $2$, $\dots$, $A_1$ contains all the elements of $M$, except $n$ of them. We deduce that the intersection $A:={\bigcap }_{k=1}^n{A}_k$ contains all the elements of $M$, except at most $1+2+\cdots +n=\frac{n(n+1)}{2}$. It follows that $$|A|\ge n^2+n+2-\frac{n(n+1)}{2}=\frac{n^2+n+4}{2}.$$ Let $A=\{x_1,x_2,\dots ,x_p\}$, with $1\le x_1<x_2<\cdots <x_p\le n^2+n+2$. Assume, by contradiction, that $A$ does not contain consecutive integers. Then $x_{i+1}\ge x_i+2$, $i=1,2,\dots ,p-1$, and we conclude that $$x_p\ge x_1+2(p-1)\ge 1+2\left(\frac{n^2+n+4}{2}-1\right)=n^2+n+3,$$ a contradiction.