Let $B$ be a set of $n$ real numbers. We choose at random an $m$-element subset $A$ of $B$. We want to know the expected value (mean) of the sum of the chosen numbers.
Let $B=\{b_1,b_2,\dots, b_n\}$. Imagine that the choosing is done sequentially. Let the random variable $X_i$ be the $i$-th number chosen, and let $X=\sum_{i=1}^m X_i.$ Then $X$ is the sum of the chosen numbers, and we want to find $E(X)$, the expected value of this sum. For any random variables $X_1, X_2, \dots, X_m$ we have $E\left(\sum_{i=1}^m X_i\right)=\sum_{i=1}^m E(X_i).$ Now we need to compute $E(X_i)$. But $X_i$ takes on the values $b_1, b_2,\dots, b_n$, each with probability $1/n$. It follows that $E(X_i)=\frac{\sum_{i=1}^n b_i}{n},$ and therefore $E(X)=\frac{m}{n}\sum_{i=1}^nb_i.$
Comments:
$1$. The OP specified that the numbers chosen are distinct. In standard terminology, this means that the choosing is done without replacement. The above argument does not depend on whether the choosing is done with replacement or without replacement: the answer under each assumption is the same.
$2$. In an answer to one of the comments, the OP seemed to say that the set $B$ consists of the first $n$ positive integers, though that wasn't entirely clear. If $B=\{1,2,\dots,n\}$, then $\sum_{i=1}^n i=\frac{n(n+1)}{2}\quad\text{and therefore}\quad E(X)=\frac{m(n+1)}{2}.$
$3$. The same argument goes through if $B$ is a multiset (some of the $b_i$ are allowed to be equal), and $A$ is a (multi)subset of $B$.
$4$. As Yuval Filmus has pointed out in a comment, we can also find the variance of $X$ by an argument that exploits the linearity of expectation.