The answer is yes, as proved by Croot [Cr03] - indeed, there are infinitely many disjoint such monochromatic solutions.

In [ErGr80] they also ask for a monochromatic representation of any $\frac{a}{b}>0$. This follows from the case of $1$ - indeed, consider the induced colouring of $\{\tfrac{n}{b}: b\mid n\}$. By the above there are $a$ solutions to\[1=\sum_i \frac{1}{n_i/b},\]and hence $a$ solutions to $\frac{1}{b}=\sum_i \frac{1}{n_i}$, where all $n_i$ are distinct (across the $a$ many solutions). Summing across all variables then yields $\frac{a}{b}=\sum_j \frac{1}{m_j}$ where all $m_j$ are distinct and the same colour, as required.

See also [298].

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