SOLVED
Let $x>0$ be a real number. For any $n\geq 1$ let $R_n(x) = \sum_{i=1}^n\frac{1}{m_i}<x$ be the maximal sum of $n$ distinct unit fractions which is $<x$.

Is it true that, for almost all $x$, for sufficiently large $n$, we have $R_{n+1}(x)=R_n(x)+\frac{1}{m},$ where $m$ is minimal such that $m$ does not appear in $R_n(x)$ and the right-hand side is $<x$? (That is, are the best underapproximations eventually always constructed in a 'greedy' fashion?)

Erdős and Graham write it is 'not difficult' to construct irrational $x$ such that this fails (although give no proof or reference, and it seems to still be an open problem to actually construct some such irrational $x$). Curtiss [Cu22] showed that this is true for $x=1$ and Erdős [Er50b] showed it is true for all $x=1/m$ with $m\geq 1$. Nathanson [Na23] has shown it is true for $x=a/b$ when $a\mid b+1$ and Chu [Ch23b] has shown it is true for a larger class of rationals; it is still unknown whether this is true for all rational $x>0$.

Without the 'eventually' condition this can fail for some rational $x$ (although Erdős [Er50b] showed it holds without the eventually for rationals of the form $1/m$). For example $R_1(\tfrac{11}{24})=\frac{1}{3}$ but $R_2(\tfrac{11}{24})=\frac{1}{4}+\frac{1}{5}.$

Kovač [Ko24b] has proved that this is false - in fact as false possible: the set of $x\in (0,\infty)$ for which the best underapproximations are eventually 'greedy' has Lebesgue measure zero. (It remains an open problem to give any explicit example of such a number.)