For $n=1$, the least possible sum is $1$. For $n=2$, the least possible sum is $4$. For $n\ge3$, the least possible sum is $\frac{n-1}{2}n!+n$.

That's not correct (you essentially made the mistake that the lowest common multiple of $1,2,...,n$ is $n!$, I think): for $n=5$, the optimum is given by $1,13,25,37,49$ with sum $125$, but your formula predicts $245$. In general, the optimal set is unique and given by $\{1+ak| 0 \leq a < n \}$ where $k = \text{lcm}(1,2,...,n-1)$, with sum $n+\frac{n(n-1)}{2} \cdot k$.

@ZetaX: Yes, that's exactly my mistake.

Let the elements of $A$ be $1\leq a_1<a_2<\cdots < a_n$, for $n\geq 3$. Let $2\leq k\leq n-1$. Consider two indices $1\leq i < j \leq n$, and a set of indices $S\subseteq \{1,2,\ldots,n\}\setminus \{i,j\}$ with $|S|=k-1$. We need $\displaystyle k\mid a_i + \sum_{\ell \in S} a_\ell$ and $\displaystyle k\mid a_j + \sum_{\ell \in S} a_\ell$, thus $k \mid a_j-a_i$. Consider $m =\operatorname{lcm}[1,2,\ldots,n-1]$; thus we need $m\mid a_j - a_i$. It is not hard to see that then the set must be $1\leq a_1 < a_1+mb_1 < a_1+mb_2 < \cdots < a_1 + mb_{n-1}$, and we also need $\displaystyle n\mid \sum_{\ell=1}^n a_\ell = na_1 + m\sum_{\ell=1}^{n-1} b_\ell$. But in order to minimize $\displaystyle \sum_{\ell=1}^n a_\ell$ we will take $a_1=1$ and $b_{\ell} = \ell$, when obviously $\displaystyle n\mid \sum_{\ell=1}^n a_\ell = n + m\dfrac {n(n-1)}{2}$, which is thus the least possible sum.

$1\leq a_1<a_2<\cdots < a_n$ $k\leq n-1$.Let's look at the number $k$.let us consider any number $k-1$ other than $a_i$ and $a_j$. let $A$ be the sum of $k-1$. $$k|a_i+A$$$$k|a_j+A$$Hence $a_i \equiv a_j (mod k)$ $$T=lcm[1,2,...,n-1]$$$$a_i=c+b_i \cdot T$$$$ \sum {a_i}= \sum c+b_i \cdot T \ge n +\frac{n(n-1)}{2}T $$for example:$a_i=(i-1)T+1$. $k\leq n-1$ easy. for $k=n$ 1)if $n$ has at least two prime divisors.exist $a$ and $b$ $(a,b)=1$,$a|T$ and $b|T$,hence $n|T$ 2)$n=p^t$ a)$p>2$ $n|\frac{n(n-1)}{2}$ b)$p=2$ $2|T$ $n|\frac{n(n-1)}{2}T$ Answer:$n +\frac{n(n-1)}{2}T $,$T=lcm[1,2,...,n-1]$

Good job bro

We will only treat the case $n \geq 3$ because the small cases are easy. Denote by $L_n$ the lcm of the numbers from $1$ to $n-1$. We will prove the answer is: $n+L_n\cdot \frac{n(n-1)}{2}$, which is obtained by the set ${1, L_n+1…(n-1)L_n+1}$ The solution is based on the following claim: Claim: For a meanly set $\mathcal{M}$ of cardinality $m$, the elements are all equal modulo $i$, $\forall 1 \leq i \leq m-1$ Proof: We will proceed by induction, the base case $m=3$ being trivial. Suppose now the claim is true for $k$ we will prove for $k+1$. Let $\mathcal{A}={a_1, a_2,…a_{k+1}}$ be a meanly set. Notice that all its subsets are meanly. Therefore ${a_1, a_2…a_k}$ and ${a_2,…a_{k+1}}$ are meanly too so by the induction hypothesis, the elements of $\mathcal{A}$ are equal mod $L_k$ Case 1: $k$ is composite. Then it may be written as the product of two coprime numbers $m$ and $n$, which are smaller than $k-1$ so we’re done $\square$ Case 2: $k$ is prime. Denote by $S$ the sum of the elements of $\mathcal{A}$. Then $S - a_i \equiv 0(\mod k)$ and $S - a_j \equiv 0(\mod k)$, therefore $a_i \equiv a_j(\mod k)$ for any two indexex $i$ and $j$, so all the elements are the same modulo $k$. And we’re done because $(L_k, k)=1$ $\square$ Write $a_i=x_i \cdot L_n + r$. If $r=0$, the sum is at least $L_n \cdot \frac{n(n+1)}{2}$ and when $r>0$ the sum is at least $n+L_n\cdot \frac{n(n-1)}{2}$, and since the second possible minimum is smaller than the first one we are done. $\blacksquare$