For a prime $$p$$, define $$f(p)$$ to be the minimum $$i$$ such that all residues mod $$p$$ can be expressed as sums of subsets of the values $$\{\tfrac{1}{1}, \tfrac{1}{2}, \tfrac{1}{3}, \dots \tfrac{1}{i} \} \bmod p$$. We need only consider residues other than $$0$$ and $$1$$, since $$0$$ is always the sum of the empty subset and $$1 = \tfrac{1}{1}$$. E.g.,

$$f(2) = 1$$, because $$1 = \tfrac{1}{1}$$.

$$f(3) = 2$$, because $$2 = \tfrac{1}{2} \bmod 3$$.

$$f(5) = 3$$, because $$2 = \tfrac{1}{3}$$, $$3 = \tfrac{1}{2}$$, and $$4 = \tfrac{1}{1} + \tfrac{1}{2} \bmod 5$$.

$$f(7) = 3$$, because $$2 = \tfrac{1}{2} + \tfrac{1}{3}$$, $$3 = \tfrac{1}{1} + \tfrac{1}{2} + \tfrac{1}{3}$$, $$4 = \tfrac{1}{2}$$, $$5 = \tfrac{1}{3}$$, and $$6 = \tfrac{1}{1} + \tfrac{1}{3} \bmod 7$$.

$$f(11) = 4$$, because $$2 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4}$$, $$3 = \tfrac{1}{4}$$, $$4 = \tfrac{1}{3}$$, $$5 = \tfrac{1}{1} + \tfrac{1}{3}$$, $$6 = \tfrac{1}{2}$$, $$7 = \tfrac{1}{1} + \tfrac{1}{2}$$, $$8 = \tfrac{1}{1} + \tfrac{1}{3} + \tfrac{1}{4}$$, $$9 = \tfrac{1}{2} + \tfrac{1}{4}$$, and $$10 = \tfrac{1}{2} + \tfrac{1}{3} \bmod 11$$.

$$f(13) = 5$$. $$2 = \tfrac{1}{2} + \tfrac{1}{5} \bmod 13$$ and $$12 = \tfrac{1}{1} + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{5} \bmod 13$$; all other residues can be expressed as sums of subsets of $$\{\tfrac{1}{1}, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4} \}$$.

How quickly does $$f(p)$$ grow with $$p$$? It must be at least $$\log_2 p$$ else there would not be enough different subsets to cover all the residues. It seems plausible to me (e.g. using an argument in which we replace each unit fraction by a random variable) that $$f(p) = \Theta(\log p)$$.

The relevance of this is for bounding the maximum denominator of an Egyptian fraction as a function of its denominator. For each $$p$$, some fraction $$x/p$$ requires a maximum denominator of at least $$f(p)p$$ in any Egyptian fraction expansion, as can be seen by considering only the terms in the expansion in which the denominator is a multiple of $$p$$. The simple lower bound for $$f(p)$$ above thus leads to a lower bound of $$\Omega(y log y)$$ on the maximum denominator in an expansion of $$x/y$$, for infinitely many pairs $$x, y$$. On the other hand, it seems plausible (although not completely obvious due to issues with repeated fractions and overlarge expansions) that something like the prime-multiples method of my Egyptian fraction code can expand any fraction $$x/y$$ using max denominator $$f(p)y$$, where $$p$$ is the prime factor of $$y$$ with the largest value of $$f(p)$$. This approach might have some hope of improving the Tenenbaumâ€“Yokota $$O(y \log^2 y / \log\log y)$$ upper bound on the max denominator.