Conjecture: If $\sum_{i=1}^n \frac{1}{x_i}=1$ then $x_i | x_j$

Question

Let $x_1,x_2,x_3,\cdots , x_n\in\mathbb{N}$ prove that if $\sum_{i=1}^n\frac{1}{x_i}=1$, then there exist $x_j,x_i$ such that $x_j | x_i$. I read that I should avoid the no clue questions, but this is a conjecture I came up with and need to know if it's true to continue my work on some project.

All I could figure out is that it's true for $n=2$ - and then $x_1=x_2=\frac{1}{2}$

Edit

The user @deif proved the conjecture if there exist atleast one $x_i$ which is prime. So to prove the conjecture entirely(if it's true) we just need to prove it with the assumption that $x_1,x_2,\cdots ,x_n$ are not primes!

Answer 1

After following Lucid's link in the comments, we have this reference, (see Example 2 in answer to Question 6), which indicates the conjecture is in fact false.

Answer 2

(This is just a partial answer, but too long for a comment).
In case there is some $x_i$ prime we proceed as it follows. Without loss of generality we assume $x_1$ is prime. Since $$1=\sum_{i=1}^n\frac{1}{x_i},$$ multiplying by $\prod_{j=1}^nx_j$ we obtain $$\prod_{j=1}^nx_j=\prod_{j=1}^nx_j\sum_{i=1}^n\frac{1}{x_i}=\sum_{i=1}^n\prod_{\substack{j=1\\ j\not=i}}^nx_j.$$ Hence, solving and taking common factor $x_1$, $$x_1(\prod_{j=2}^nx_j-\sum_{i=2}^n\prod_{\substack{j=2\\ j\not=i}}^nx_j)=\prod_{j=2}^nx_j.$$ Then, since $x_1$ is prime it must divide some $x_j$, where $2\leq j\leq n$.