$\frac{1}{n}$ as a difference of Egyptian fractions with all denominators $<n$
Partial answer. It shows for example that if $N$ is large enough and satisfies $P^*(N)^2\leq N$ then $N\in S$. (I think that the methods used in the paper are powerful enough to answer the question completely, In Progress)
The condition that the exceptions are tiny multiple of prime powers reminded me of the studies of representing the unity as the sum of Egyptian fractions by P. Erdös, E. S. Croot, and G. Martins.
After some google search, I found the following paper by Greg Martin, Denser Egyptian fractions, Acta Arith.
Let for any $y>1$ a real number, $\pi^*(y)$ denote the number of prime powers less than $y$, and for any positive integer $n$, $P^∗(n)$ denote the largest prime power dividing $n$.
In the above paper, it is proven in proposition 7 that,
Proposition 7. Let $y$ be a sufficiently large real number, and let $\frac{a}{b}$ be a rational number satisfying $\frac{a}{\log(y)} < \frac{a}{b} < 1 $ and $P^∗(b) \leq y$. Then there is a set $S$ of integers satisfying:
- (i) $S$ is contained in $[1, 2y^4]$;
- (ii) $|S| = 2π^∗(y)$;
- (iii) $a/b = \sum_{s\in S} \frac 1 s$
After reviewing the proof of this proposition, if we don't care about the size, then we can actually prove the following (already contained in the proof of proposition 7).
Proposition : Let $y$ be a sufficiently large real number, and let $\frac{a}{b}$ be a rational number satisfying $\frac{a}{\log(y)} < \frac{a}{b} < 1 $ and $P^∗(b) \leq y$. Then there is a set $S$ of integers satisfying:
- (i) $S$ is contained in $[1, y^2]$;
- (iii) $a/b = \sum_{s\in S} \frac 1 s$
As a corollary of this proposition, one can prove that for any large number $N$ such that $P^*(N)^2\leq N$ there exists a set $S\subset [1,N-1]$ such that : $$\frac 1 n =(\frac 12+\frac 13 +\frac 1 6)- \sum_{s\in S} \frac 1 s $$
By taking $\frac a b = \frac {n-1} n $, $y=\sqrt n $.