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Necessity: we search for the second smallest arithmetic mean, the smalles being $1$ for the subset $\{1\}$. Of course, the subset $A$ with the second smallest arithmetic mean has to contain $1$ so let $A = \{1\} \cup B$ with $B \subset \{2,3, \dots ,n\}$ and $|B| = t \neq 0$. We have $m_{A} = \frac{1 + m_{B}t}{t} \ge \frac{1 + 2t}{t} \ge \frac{3}{2}$. And so, since $\{1\}$ is connected to a subset, we have $k\ge \frac{1}{2}$. Sufficiency: we have that $1 \le m_{M} \le n$ for any subset $M$ of $\{1,2,\dots,n\}$. Consider the subsets $\{1\}, \{1,2\}, \{2\}, \dots , \{n-1,n\}, \{n\}$. Their arithmetic means are equal to $1, \frac{3}{2}, 2, \dots , \frac{2n-1}{2}, n$ and hence these form a chain, but also they "cover" the whole interval $[1,n]$ so every other subset is connected to some vertex of this chain so we are done.