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we prove all $k>2018$ are the answer. for $k<2019$ just look at the closed interval of $[2019,2019+\frac{1}{1000000000}]$ for $k>2020$ let : $\frac{1}{a_1}+...+\frac{1}{a_n} \in (a,b)$ then just add $(k-2020)\frac{1}{N}$ where $N$ is a big enough integer . so we are done there aswell. the main part seems to be $k=2019$ we prove it works. let $b_i$ be a sequance of sets of cardinality of $2020$ with positive integers as elements such that $$\sum_{x \in b_i}{\frac{1}{x}} \in [a,b]$$. now consider all of them in decreasing manner and let $$b_i =\{a_{i,1},a_{i,2},...,a_{i,2020}\}$$with $a_{i,1} \le a_{i,2} \le ... \le a_{i,2020}$ now take the least integer $j$ such that the set $c_j=\{a_{1,j},a_{2,j},...\}$ is not bounded. such set exists clearly. and we have $c_j,...,c_{2020}$ are all un_bounded. now look at $a_{i,1},...,a_{i,j-1}$ there are only finitly many possible such sets. so using pigeon hole prinsible and the fact that there are infinitly many of them. we may assume that $a_{i,1},...,a_{i,j-1}$ are some constants now for $k=2019$ take the set $T=\{a_{i,1},...,a_{i,j-1},N,N,...,N\}$ such that $N$ has appeard $2020-j$ times so the sum $\sum_{x \in T} $ needs to be in the closed inteval of $[a,b]$. let $W=\{1,2,...,2020\}$ consider the sums $A_t=\sum_{i \in W}{\frac{1}{a_{t,i}}}$ let $\frac{1}{a_{i,1}}+...+\frac{1}{a_{i,j-1}}=C$ and $A_t=B_t+C$ . our sum is $M=C+\frac{2020-j}{N}$ take $N$ big enough such that $M=C+\frac{2020-j}{N} \le A_l$ for some $l$ now we need $C+\frac{2020-j}{N} \ge A_r$ for some $r$ this is also easy since $a_{r,j},...,a_{r,2020}$ for big enough $r$ are all bigger than $N$. so $k>2018$ works ($k=2020$ is trivial )

Answer: $\boxed{ k \geq 2019}$. Note that any $n-good$ number is also an $(n+1)-good$ number; letting an $n-good \ number$ to be $\frac{1}{a_1} + \frac{1}{a_2}+\cdots+\frac{1}{a_n}$, it is also representable as $\frac{1}{a_1} + \frac{1}{a_2}+\cdots+\frac{1}{2a_n}+\frac{1}{2a_n}$. From therein we get that any $k \geq 2020$ is viable, since we can just extend an arbitrary $2020-good$ in the interval $[a,b]$ to infinity-good ones. However, notice that any $a-good$ number is at most $a = \dfrac{1}{1}+ \ldots+\dfrac{1}{1}$ added $a$ times. By considering the interval $[a+\epsilon< a+\dfrac{1}{2},2020]$ where the $2020-good$ numbers $x_n = $ $1+1+\ldots+1+\dfrac{1}{n} = 2019+\dfrac{1}{n}$ lies, we get that $a \leq 2018$ is not viable. It remains to check $k = 2019$. ${\color{green}\rule{25cm}{3pt}}$ Suppose we have an interval $[a,b]$ where the numbers $x_n = \dfrac{1}{a_1^{(n)}}+\dfrac{1}{a_2^{(n)}}+\cdots+\dfrac{1}{a_{2020}^{(n)}} $ lies. For each number, we can WLOG assume that $a_i^{(n)} \leq a_j^{(n)}$ when $i \leq j$. If all $a_i^{(n)}$ is unbounded, $a = 0$, and there is easily infinitely many $2019-good$ numbers near $0$. Otherwise, since there are infinitely many of those $x_n$, there must be a group of indices $\{i,i+1,\ldots, 2020\}$ so $a_j^{(n)}, i\leq j\leq 2020$ unbounded when $n$ is flung towards $+\infty$ (note: the group must be in that form, as if the index $k$ is unbounded but $k+1$ isn't, then there exists two numbers $+\infty = a_k^{(n)} \leq a_{k+1}^{(n)} \leq C$.) Because the possibilities of $(a_j^{(n)}, 1 \leq j \leq i-1)$ is bounded even when $n$ is flung towards $+\infty$, by infinite $PHP$ on the $(i-1)-tuples$ $(a_1^{(n)}, a_2^{(n)}, \ldots, a_{i-1}^{(n)})$, we can force the infinite existence of one $(i-1)-tuple$. Let that tuple be $(b_1,b_2,\ldots,b_{i-1}).$ Then, we can infer that there exists numbers $x_n = \dfrac{1}{b_1}+\cdots+\dfrac{1}{b_{i-1}} + \dfrac{1}{\text{big}}+ \cdots+ \dfrac{1}{\text{big}}$ in the interval $[a,b]$ arbitrarily close to $\dfrac{1}{b_1}+\cdots+\dfrac{1}{b_{i-1}}$. This is obtained when we pick $x_n$ with $a_j^{(n)} = b_j, 1\leq j \leq i-1$. Then, we can infer that the $(i-1)-good$ number $\dfrac{1}{b_1}+\cdots+\dfrac{1}{b_{i-1}}$ is in the interval $[a,b]$, due to continuity. We are done.