Brazilian Mathematical Olympiad

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,
$$\begin{align*} \sum_{k=1}^{n-1} (n-k) \cdot 2^{k-1} &= 2^n - n - 1 \\ \iff \sum_{k=1}^{n-1} (n-k) \cdot 2^k &= 2^{n+1} - 2n - 2 \\ \iff \sum_{k=0}^{n-1} (n-k) \cdot 2^k &= 2^{n+1} - 2n - 2 + n = 2^{n+1} - n - 2 \end{align*}$$
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, $M+m = (n+1)(2^n-1) \iff M-m = (n+1)(2^n-1)-2(2^{n+1}-n-2) = (n-3) \cdot 2^n + n + 3$.