55rd Ukrainian National Mathematical Olympiad - Third Round

If we consider two subcommittees, they have exactly one leader in common, therefore, together they have exactly 5 members. Hence there are at least 5 people in the committee. Denote them by $A=\{1,2,3,4,5\}$. However it's impossible to choose leaders for another subcommittee out of them. Therefore, the committee must have at least 6 members. Here's an example of four subcommittees and their leaders: $$A=\{1,2,3,4,5,6\}\to \{1,2,3\};\{3,4,5\};\{1,5,6\};\{2,4,6\}.$$