Let's start from the beginning. about a week ago, I asked a question in MSE with(almost) the same title.
- Here is the link to said question - If each term in a sum of positive integers divides that sum, then there must be one term that divides another.
This claim turned out to be false, due to a counter example that disproves an equivalent conjecture! - Here is the equivalent conjecture - Conjecture: If $\sum_{i=1}^n \frac{1}{x_i}=1$ then $x_i | x_j$
- Here is the article contains the counter example(the counter example is in example $2$ after question $6$) - https://www.sciencedirect.com/science/article/pii/0012365X73901362
After looking again at everything(the conjecture, the article, the different questions and answers, and more), I've come to the conclusion that for what I'm trying to do, it is enough to prove the conjecture with the condition that each two terms in the finite sequence are NOT co-prime.
i.e $\gcd(x_j,x_l)>1$ for every $1\le j\neq l\le n$
Let $\begin{align} \{ x_i \}_{i=1}^n \end{align}$ be a finite sequence such that $x_i\in\mathbb{N}$ that satisfies $\gcd(x_j,x_l)>1$ for every $1\le j\neq l \le n$.
Prove that if $x_i$ divides $\begin{align}\sum_{i=1}^n x_i\end{align}$, $\forall{ 1\le i\le n}$ then there are $1\le h\neq k\le n$ such that $x_h$ divides $x_k$.
At my previous question, I've tried to prove the statement by induction, we managed to prove the base cases, for $n=2,3$. Obviously, those still hold for this version of the question, because it is a weaker variant.
EDIT -
After it was brought to my attention, this claim is also false.
But a new claim that can be interesting is if that is true for $\gcd(x_1,x_2,\cdots x_n)=1$