A finite set of positive integers $A$ is called [i]meanly[/i] if for each of its nonempy 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 $n$, determine the least possible sum of the elements of a meanly $n$-element set.
To determine the least possible sum of the elements of a meanly \( n \)-element set \( A \), we need to ensure that for each non-empty subset of \( A \), the arithmetic mean of its elements is a positive integer. 

1. **Understanding the Condition**:
   Let \( A = \{a_1, a_2, \ldots, a_n\} \) with \( 1 \leq a_1 < a_2 < \cdots < a_n \). For any subset \( \{a_{i_1}, a_{i_2}, \ldots, a_{i_k}\} \) of \( A \), the arithmetic mean \(\frac{1}{k}(a_{i_1} + a_{i_2} + \cdots + a_{i_k})\) must be an integer. This implies that the sum \( a_{i_1} + a_{i_2} + \cdots + a_{i_k} \) must be divisible by \( k \).

2. **Divisibility Condition**:
   Consider any \( k \) distinct elements \( a_{i_1}, a_{i_2}, \ldots, a_{i_k} \) from \( A \). We need \( k \) to divide the sum of these \( k \) elements. This implies that for any \( k \leq n \), the sum of any \( k \) elements of \( A \) must be divisible by \( k \).

3. **Modular Arithmetic**:
   Let \( A \) be the sum of \( k-1 \) elements other than \( a_i \) and \( a_j \). We have:
   \[
   k \mid (a_i + A) \quad \text{and} \quad k \mid (a_j + A)
   \]
   This implies:
   \[
   a_i \equiv a_j \pmod{k}
   \]
   Hence, all elements \( a_i \) must be congruent modulo \( k \).

4. **Least Common Multiple**:
   To satisfy the condition for all \( k \leq n-1 \), we need:
   \[
   T = \text{lcm}(1, 2, \ldots, n-1)
   \]
   Let \( a_i = c + b_i \cdot T \) for some integer \( c \) and distinct integers \( b_i \). The sum of the elements of \( A \) is:
   \[
   \sum_{i=1}^n a_i = \sum_{i=1}^n (c + b_i \cdot T) = nc + T \sum_{i=1}^n b_i
   \]

5. **Minimizing the Sum**:
   To minimize the sum, we can choose \( c = 1 \) and \( b_i = i-1 \). Thus:
   \[
   a_i = 1 + (i-1)T
   \]
   The sum of the elements is:
   \[
   \sum_{i=1}^n a_i = \sum_{i=1}^n (1 + (i-1)T) = n + T \sum_{i=0}^{n-1} i = n + T \cdot \frac{n(n-1)}{2}
   \]

6. **Special Cases**:
   - If \( n \) has at least two prime divisors, then \( n \mid T \).
   - If \( n = p^t \) for some prime \( p \):
     - If \( p > 2 \), then \( n \mid \frac{n(n-1)}{2} \).
     - If \( p = 2 \), then \( 2 \mid T \) and \( n \mid \frac{n(n-1)}{2}T \).

Thus, the least possible sum of the elements of a meanly \( n \)-element set is:
\[
n + \frac{n(n-1)}{2}T
\]
where \( T = \text{lcm}(1, 2, \ldots, n-1) \).

The final answer is \( \boxed{ n + \frac{n(n-1)}{2} \cdot \text{lcm}(1, 2, \ldots, n-1) } \).