Summing max elements of all subsets.

Let $S=\{1,2,\ldots,N\}$ . For any $n<N$ , the number of subsets of $S$ for which $n$ is the maximal element is $2^{n-1}$ , since each natural number less than $n$ is either in the subset, or it isn't. So $1$ will occur $1$ time in the computation of $f(N)$ , $2$ will occur $2$ times, $3$ will occur $4$ times, etc. Thus $f(19) = \sum_{n=1}^{19} n2^{n-1} = \boxed{9437185}$

We'll try to find a recursive relation for $f(n)$ .
Note that for a natural $n$ added to the set $S = \{1,2,3 \dots n-1\}$ , there are exactly $2^{n-1}$ new sets created for which $n$ is the maximal element. So, we can say that: $f(n) = f(n-1) + n2^{n-1}$ This recursion can be solved using a program or we can find a closed form to it.
We can see that since $f(1)=1$ $f(n) = \sum_{i=1}^{n} i2^{i-1}$ Consider, $g(n,a) = \sum_{i=1}^{n}a^i = \dfrac{a(a^n -1)}{a-1}$ for some real $a$ and $a \neq 1$ Differentiating w.r.t $a$ $\frac{\partial g(n, a)}{\partial a} = g'(n, a) = \sum_{i=1}^{n}ia^{i-1} = \dfrac{(a-1)((n+1)a^n-1) - (a^{n+1} -a)}{(a-1)^2}$ We see that, $g'(n, 2) = f(n) = \dfrac{(n+1)2^n -1 - 2^{n+1} +2}{1} = (n+1)2^n - 2^{n+1} +1$ Plugging in $n=19$ $f(19) = 9437185$