XXIV OBM

Let $m$ and $M$ be the sum of the minima and maxima of all subsets. Since the diameter of a set is the difference between its maximum and its minimum, the desired sum is $M-m$. We may include unitary subsets, since their minima and maxima coincide.

The number $k$, $1\le k\le n$, is the minimum of all subsets of the form $\{k\}\cup A$, where $A\subset \{k+1,k+2,\dots ,n\}$. So $k$ is the minimum of $2^{n-k}$ subsets. Hence $$m=\sum _{k=1}^{n}k\cdot 2^{n-k}=\sum _{k=0}^{n-1}(n-k)\cdot 2^k.$$ Now we count the number of subsets with diameter $k$. Let $a$ be the minimum of a subset. The maximum is $a+k$. Since $a+k\le n$, one may choose $a$ in $n-k$ ways. Since there are $k-1$ numbers between $a$ and $a+k$, there are $(n-k)\cdot 2^{k-1}$ subsets with diameters $k$. Since there are $2^n-n-1$ non-empty and non-unitary subsets, $$\begin{array}{rl}\sum _{k=1}^{n-1}(n-k)\cdot 2^{k-1}& =2^n-n-1\\ ⟺\sum _{k=1}^{n-1}(n-k)\cdot 2^k& =2^{n+1}-2n-2\\ ⟺\sum _{k=0}^{n-1}(n-k)\cdot 2^k& =2^{n+1}-2n-2+n=2^{n+1}-n-2\end{array}$$ and $m=2^{n+1}-n-2$.

In order to compute $M$, notice that the association $A=\{a_1,a_2,\dots ,a_m\}\to f(A)=\{n+1-a_1,n+1-a_2,\dots ,n+1-a_m\}$ is clearly a bijection that transform minima $\ell$ in maxima $n+1-\ell$. Since there are $2^n-1$ non-empty subsets, $M+m=(n+1)(2^n-1)⟺M-m=(n+1)(2^n-1)-2(2^{n+1}-n-2)=(n-3)\cdot 2^n+n+3$.