Let $a_i,i \geq 2 = (i-2)!$ and $a_1=n!$ Then can be proved by induction that $\frac{1}{2a_2}+\frac{1}{3a_3}+...+\frac{1}{na_n}=\sum_{t=2}^n \frac{1}{t(t-2)!} = 1-\frac{1}{n!}$ So $\dfrac{1}{a_1}+\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\dots+\dfrac{1}{na_{n}}=1$

Use $\frac{1}{n} = \frac{n+1}{(n+1)n} = \frac{n}{(n+1)n} + \frac{1}{(n+1)n} = \frac{1}{n+1} + \frac{1}{(n+1)n}$ and induct.

If $\frac{1}{b_{1}}$ + $\frac{1}{ 2b_{ 2}}$ + ... + $\frac{1}{ nb_n}$ = 1 take $a_{1}$=$b_{1} $ $ a_{ 2}$= $ b_{ 2}$ ... $a_{n-1}$= $b_{n-1}$ $ a_n$= ($n+1$)$ b_n$ $ a_{ n+1}$= $ b_{ n}$ then: $\frac{1}{a_{1}}$ + $\frac{1}{ 2a_{ 2}}$ + ... + $\frac{1}{ na_n}$ + $\frac{1}{ (n+1)a_{ n+1}}$= 1

inducc Claim: For any $n$, we can find $a_k$ for $1 \le k \le n$ such that $\sum_{k=1}^n \frac{1}{ka_k} = \frac{n}{n+1}$. Proof: Base case for $n = 1$ works since $a_1 = 2$ gives $\frac{1}{2} = \frac{1}{2}$. For the inductive step, we let $a_{n+1} = n+2$ and note that; $$ \sum_{k=1}^{n+1} \frac{1}{ka_k} = \frac{n}{n+1} + \frac{1}{(n+2)(n+1)} = \frac{n+1}{n+2}. \Box$$ Now for any $n$, we can let $a_n = 1$ and find $a_i$ with $1 \le i \le n-1$ such that $\sum_{k=1}^{n-1} \frac{1}{ka_k} = \frac{n-1}{n},$ so $\sum_{k=1}^{n} \frac{1}{ka_k} = \frac{n-1}{n} + \frac{1}{n} = 1.$

Could be done easily by the induction.

Am I doing something wrong, or are the above solutions unnecessarily complicated? Just take $(a_1,a_2,a_3,\dots, a_{2020}) = (2020,1,2,3,\dots, 2019)$, and it equals $1$ by basic telescoping.

We prove the more general statement that there exists positive integers $a_1,a_2,\dots,a_n$ such that \[\frac{1}{a_1}+\frac{1}{2a_2}+\dots + \frac{1}{na_n}=1\]with $a_n=1$. First note that when $n=3$, $a_1=6$,$a_2=1$ and $a_3=1$ is such a triple. Then, say we have positive integers $a_1,\dots,a_k$ such that \[\frac{1}{a_1}+\frac{1}{2a_2}+\dots + \frac{1}{ka_k}=1\]with $a_k=1$ for some $k\geq 3$. Then, note that, \[\frac{1}{ka_k}=\frac{1}{k}=\frac{1}{k(k+1)}+\frac{1}{k+1}\]Thus, we have a solution for $k+1$ as, $(a_1,a_2,\dots,a_{k-1},k+1,1)$. This solution also has $a_{k+1}=1$ and thus, our induction is complete and indeed there exists $a_1,a_2,\dots,a_n$ such that \[\frac{1}{a_1}+\frac{1}{2a_2}+\dots + \frac{1}{na_n}=1\]for all $n\geq 3$.

we only have to find a set of $a_{n} $ that satisfies. ${a_1}=2$, $a_{2020}=1$, $a_n=n+1$ for $2 \le n \le 2019$. done.