For $n=1$ we can take any natural $x_1\geq 1$, since $\dfrac {1}{x_1} - \dfrac {1}{x_1} = 0$. For $n=2$ we can take $x_1=1$ and any natural $x_2>1$, since $\dfrac {1}{1} + \dfrac {1}{x_2} - \dfrac {1}{1\cdot x_2} = 1$. For $n>2$, it is reminiscent of a classical construction used with Egyptian fractions. Using the identity $\dfrac {1}{a} = \dfrac {1}{a+1} + \dfrac {1}{a(a+1)}$, we can start with $x_1 = 2$, $\displaystyle x_k = 1+ \prod_{i=1}^{k-1} x_i$ for $2\leq k \leq n-1$ and $\displaystyle x_n = \prod_{i=1}^{n-1} x_i$ to get $\displaystyle \sum_{i=1}^n \dfrac {1}{x_i} = 1$. Only that for our purposes we will take $\displaystyle x_n = -1 + \prod_{i=1}^{n-1} x_i$ to get $\displaystyle \sum_{i=1}^n \dfrac {1}{x_i} - \dfrac {1} {\prod\limits_{i=1}^n x_i} = 1$, since $\dfrac {1} {x_n} - \dfrac {1} {\prod\limits_{i=1}^n x_i} = \dfrac {1} {x_n} - \dfrac {1}{x_n(x_n+1)} = \dfrac {1}{x_n+1} = \dfrac {1} {\prod\limits_{i=1}^{n-1} x_i}$. Thus we proved even more, namely the expression amounts to $1$ for any $n>1$.

My solution: Lemma: for every $n$ there are $n$ natural numbers $x_1<x_2<\cdots <x_n$ such that for every $1\le i\le n$: $x_i\mid x_1x_2\cdots x_{i-1}x_{i+1}x_{i+2}\cdots x_n+1$. Prove: we prove it by induction. Assume that it is true for $n$ and $x_1<x_2<\cdots <x_n$ satisfy the conditions. Then take $x_{n+1}=x_1x_2\cdots x_n+1$ then it's easy to see that $x_1,x_2,\cdots ,x_{n+1}$ satisfy the conditions of the lemma for $n+1$. Back to the main problem: Note that $\frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n}-\frac{1}{x_1x_2\cdots x_n}=\frac{x_2x_3\cdots x_n+x_1x_3\cdots x_n+\cdots+ x_1x_2\cdots x_{n-1}}{x_1x_2\cdots x_n}\in \mathbb{N}\longrightarrow x_i\mid x_1x_2\cdots x_{i-1}x_{i+1}x_{i+2}\cdots x_n-1$ so it is enough to prove that there are $n$ distinct positive integers $x_1<x_2<\cdots <x_n$ such that for every $1\le i\le n$: $x_i\mid x_1x_2\cdots x_{i-1}x_{i+1}x_{i+2}\cdots x_n-1$ we prove this by induction. By the lemma there are $n$ positive integers such that each of them divides the product of the others plus one. Take these $n$ numbers. then take $x_{n+1}=x_1x_2\cdots x_n-1$ then it is easy to see that these $n+1$ numbers satisfy the condition of the main problem.

it is easy to prove that there are $ n $ numbers that satisfy the following condition. $$\frac{1}{y_1}+...+\frac{1}{y_n}+ \frac{1}{y_1...y_n} = 1$$$x_1 =y_1,...,x_n=y_n,x_{n+1}=y_1y_2...y_n-1$.