Belorusija 2012

(Solution of A. Zhuk.) First, if $M=\{1,2,...,6\}$, then due to the conditions i) and ii) we have $|M_1|+|M_2|+...+|M_6|=21$. It follows that $|M_k|\ge 4$ for some index $k$. There is nothing to prove if $|M_k|=6$. If $|M_k|=5$, then there exists $l\notin M_k$, and $M_k\cup M_l=M$. If $|M_k|=4$, then there exist distinct $l$, $m\notin M_k$, and either $M_k\cup M_l=M$ or $M_k\cup M_m=M$. It remains to show that $n=7$ does not satisfy the problem condition. Indeed, for $M=\{1,2,...,7\}$, the sets $M_1=\{1,2,3,4\}$, $M_2=\{2,4,6,7\}$, $M_3=\{2,3,5,7\}$, $M_4=\{3,4,5,6\}$, $M_5=\{1,2,5,6\}$, $M_6=\{1,3,6,7\}$, $M_1=\{1,4,5,7\}$ give the counterexample needed.