Yes. Take a large $N$. Let $E := \{n \le N : p \mid n \implies p \le \frac{N}{100}\}$, where $p$ stands for a prime.
Note that $$\biggl|[N]\setminus E\biggr| \le \sum_{\frac{N}{100} \le p < N} \frac{N}{p} \le C\frac{N}{\log N},$$ implying \begin{equation}\tag{1} |E| \ge \Bigl(1-o(1)\Bigr) N.\end{equation}
Now, $$\log \text{lcm}(E) = \log \prod_{p \le \frac{N}{100}} p^{\lfloor \log_p N \rfloor} \le \sum_{p \le \frac{N}{100}} \log N \le \frac{N}{10}.$$
In other words, \begin{equation}\tag{2} \text{lcm}(E) \le e^{N/10} \le 2^{N/5}.\end{equation}
Therefore, for any $A \subseteq E$, we have $\sum_{n \in A} \frac{1}{n} = \frac{p}{\text{lcm}(E)}$ for some $p \le N \, \text{lcm}(E) \le 2^{N/4}$.
Finishing up, each $A \subseteq E$ gives rise to a pair $(\sum_{n \in A} n, \text{lcm}(E)\sum_{n \in A} \frac{1}{n})$, which lives in $[N^2] \times [2^{N/4}]$, the size of which is at most $2^{N/3}$. Since there are $2^{|E|} \ge 2^{N/2}$ many subsets of $E$, the pigeonhole principle guarantees two (distinct) with the same pair.