Q:

Minimum sum of set whose average of subsets is positive integer

A finite set of positive integers $A$ is called meanly if for each of its nonempty subsets the arithmetic mean of its elements is also a positive integer. In other words, $A$ is meanly if $\frac{1}{k}(a_1 + \dots + a_k)$ is an integer whenever $k \ge 1$ and $a_1, \dots, a_k \in A$ are distinct.
Given a positive integer $n$, what is the least possible sum of the elements of a meanly $n$-element set?
Source: Middle European Math Competition

A:

For motivation, let's do some small ones by hand.  For $n=1$, the set can be $\{1\}$ with sum $1$.  For $n=2$, we can put $1$ in, but then we can't have $2$, which would give mean $\frac 32$ for the whole set.  But $\{1,3\}$ with sum $4$ works.  It should be clear that if $n \ge 2$ all the numbers have to have the same parity.  For $n=3$, then, we can't use $4$, but $\{1,3,5\}$ with sum $9$ works.  But if $n\ge 4$, we can't have all of $1,3,5$, as whatever the fourth element $k$ is, either $\{1,3,k\}$ or $\{1,5,k\}$ will not have an integral mean.  The best we can do is $\{1,7,13,19\}$ with sum $40$  
This indicates that all the numbers in the set must be equivalent $\pmod {\operatorname{lcm}(1,2,3,\dots n-1)}$  Let $p=\operatorname{lcm}(1,2,3,\dots n-1)$, then our set is $\{1,1+p,1+2p,\dots 1+(n-1)p\}$ with sum $\frac 12np(n-1)+n$

