$m_i, n_j$ integers and $\{m_i\}_{i=1}^{k}\neq\{n_j\}_{j=1}^{k'}.$ Does $\sum\frac{1}{m_i}=\sum\frac{1}{n_j}\implies\sum m_i\neq\sum n_j?$

Question

Suppose $\{m_i\}_{i=1}^{k}$ and $\{n_j\}_{j=1}^{k'}$ are each finite subsets of $\mathbb{N},$ $\{m_i\}_{i=1}^{k}\neq\{n_j\}_{j=1}^{k'},$ and $\displaystyle\sum_{i=1}^{i=k}\frac{1}{m_i} = \sum_{j=1}^{j=k'} \frac{1}{n_j}. $ Is it true then that $\displaystyle\sum_{i=1}^{i=k}m_i \neq \sum_{j=1}^{j=k'} n_j\ ? $

The contrapositive statement seems similar in difficulty.

Answer 1

Why, of course not.

$$ \frac13+\frac1{12}+\frac1{21}+\frac1{28}=\frac14+\frac16+\frac1{18}+\frac1{36}=\frac12$$ while $$ 3+12+21+28=4+6+18+36=64 $$