Egyptian fractions with very large denominators

First, can you establish that the algorithm always halts for this problem?

The reason I ask, is that several years ago, it was (and I believe still is) an open problem that if $\frac{a}{b}$ is a rational number with $b$ odd---

Can $\frac{a}{b}$ be expressed as $\sum \frac{1}{x_{i}}$, where the $x_{i}$ are chosen to be the least possible odd integers that leave a nonnegative remainder, in a finite number of terms?

This problem received a fair amount of attention from Richard K. Guy in his Unsolved Problems in Number Theory, 3rd Edition. In it, he highlights some of the efforts by certain individuals, such as Stan Wagon, who has tackled the fraction $\frac{3}{179}$, which produced 19 terms, the last of which has 439492 digits. This, Guy indicates was beaten by David Baily who considered $\frac{3}{2879}$ and used the greedy algorithm to expand it to the sum of 21 unit fractions, the last term of which has 3018195 digits. Later, it was shown that $\frac{5}{5809}$ has at least 22 terms with the last being in excess of 60,000,000 digits.

Incidentally, although the 3rd edition of Guy's book is the most recent, it was however, published in 2005. Nevertheless, it seems that the problem is still open. See, for instance:

https://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/203587#203587

I hope this puts a little bit of light on the problem.

There's a very simple answer to the original question. The greedy algorithm, applied to $a/b=2/(2n-1)$, gives $${2\over2n-1}={1\over n}+{1\over n(2n-1)}$$ so in the notation of the original question we have $${D\over b}={n(2n-1)\over2n-1}=n$$ which can be made arbitrarily large.