$ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}>0$ and $\dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}\in\mathbb{N}\Longrightarrow $ $\Longrightarrow \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}\ge1$. A solution for equality is $a_k=nk, \forall k\in\{1,2,\dots,n\}$. Otherwise, $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}\le$ $\le \dfrac{1}{1} +\dfrac{2}{1} +\cdots +\dfrac{n}{1}=\dfrac{n(n+1)}{2}$. Results: $1\le \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}\le\dfrac{n(n+1)}{2}$. For a fixed $n\in\mathbb{N}$, let the statement $P_n$: $\forall m\in\mathbb{N}, 1\le m\le\dfrac{n(n+1)}{2}, \exists a_1, a_2,\dots, a_n\in\mathbb{N}$ such that $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}=m$. We will prove that $P_n$ is true for all $n\in\mathbb{N}$. Proof by induction: Step 1: $n=1$. $\exists a_1=1$ such that $1\le \dfrac{1}{a_1}\le m, \forall 1\le m\le \dfrac{1\cdot2}{2}=1$. Step 2: Assume $P_n$ is true for $n\ge 1$. Then $P_{n+1}$ is true. 2.a) for $m=1$: let $a_k=(n+1)k, \forall k\in\{1,2,\dots,n, n+1\}$. $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}+\dfrac{n+1}{a_{n+1}}=1$. 2.b) Let $a_1, a_2,\dots, a_n\in\mathbb{N}$ such that $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}=m$, where $1\le m\le \dfrac{n(n+1)}{2}$. For $a_{n+1}=n+1$ results $\dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}+\dfrac{n+1}{a_{n+1}}=m+1; 2\le m+1\le \dfrac{n(n+1)}{2}+1$. 2.c) Let $a_1, a_2,\dots, a_n\in\mathbb{N}$ such that $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}=m$, where $1\le m\le \dfrac{n(n+1)}{2}$. For $a_{n+1}=1$ results $\dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}+\dfrac{n+1}{a_{n+1}}=m+n+1; n+2\le m+n+1\le \dfrac{(n+1)(n+2)}{2}$. From $2.a), 2.b), 2.c)$ results: $\forall M\in\mathbb{N}, M\in\{1\}\cup\left[2,\dfrac{n(n+1)}{2}+1\right]\cup\left[n+2,\dfrac{(n+1)(n+2)}{2}\right]$, $\exists a_1, a_2, \dots, a_n, a_{n+1}$ such that $\dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}+\dfrac{n+1}{a_{n+1}}=M$. But for $n\ge1$ holds $n+2\le \left(\dfrac{n(n+1)}{2}+1\right)+1$. Results: $\forall M\in\mathbb{N}, 1\le M\le \dfrac{(n+1)(n+2)}{2}, \exists a_1, a_2, \dots, a_n, a_{n+1}$ such that $\dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n}+\dfrac{n+1}{a_{n+1}}=M$, hence $P_{n+1}$ is true. $\textbf{Final answer}$: The set of the integer numbers that can be written as $ \dfrac{1}{a_1} +\dfrac{2}{a_2} +\cdots +\dfrac{n}{a_n} , $where $ a_1,a_2,...,a_n $ are natural numbers, is: $$S_n=\left\lbrace1,2,\dots,\dfrac{n(n+1)}{2}\right\rbrace$$.