Thanks to Jeremy Rouse, I have developed some heuristics and a nomogram-like approach to computing (bounds on a candidate for) the next smallest $n$ which admits an exact packing.

I call them heuristics, but they are modest extensions of known facts which should be easy to prove. The primary observation is that a prime power cannot occur just once in an integral packing, and if it occurs more than once, it must occur in a cancellative combination. Thus $1/25$ or $1/27$ can't be the only denominators in an integral packing with that high a power of $5$ or $3$, nor can we just have both of $1/25$ and $1/50$, but we can have $1/27$ and $1/54$ together, and we can have $1/50$ and $1/75$ together, or we can have any of the above with enough high powers of the same prime. This follows from looking at the $p$-adic value of $a/b + c/(dp^n)$ for $c$ and $d$ coprime to $p$ and $p^n$ not dividing $b$.

To help in determining which $n$ are feasible for an exact packing, I arrange the positive integers in several rows, with the $i$th row containing those numbers that are both multiples of the $i$th prime and have that prime as its largest prime factor. Part of the table looks like this with prime powers marked:

```
2 4* 8* 16*
3 6 9* 12 18 24 27*
5 10 15 20 25* 30
7 14 21 28
```

I then use it to find a good subset of denominators, namely a set of integers at most $n$ whose
reciprocal sum is $B_n - 1$. The bonus is that if this subset contains another which helps form an exact packing into $B_n - 2$ bins, then this good subset helps form an exact packing for $n$.

I can use this to find Jeremy Rouse's solution pretty quickly. As $H_{11}$ is near enough to $3 +1/11$, I first determine unfeasible $n\gt 11$ by noting that if $n\lt 18$, then $9$ cannot be part of a good subset,if $n \lt 24$ this excludes $8$, and if $n\lt 28$, then $7$ and its small multiples are excluded. Similarly, multiples of primes greater than 10 and prime powers greater than $10$ are excluded.

This leads to the conclusion that for $n\lt 18$ the sum of reciprocals of allowed numbers is less than $3$, and exceeds $3$ by $1/24$ only when $n=20$. When $n=24$ we get a potential good subset which excludes $12$ (because $12$ is needed to handle the excess) and Jeremy's solution pops out.

I can use the table also to determine quickly that after $30$, the next feasible number to admit an exact packing must be greater than $51$, primarily by balancing potential members greater than $31$ of a good subset against members less than $31$ which can't be part of a good subset for $n$: if the smaller denominators overpower the larger ones, then that $n$ is not feasible.

I then group the potential members by line, seeing if I can find subsets of each line which may be part of a good subset. Leaving out those found by Jeremy, for $n=66$ I find $16$ (contributing $1/16$), $12,27,36,48,54 (3/16)$, $30,60 (1/20),$ $7$ through $42 (7/20)$, $11,22,33,55,66 (1/5)$, and $13,26,52,65 (3/20)$, which gives an exact packing for $n=66$.

I have not determined if $65,63,$ or $60$ are feasible, but I suspect not as multiples of $17$ and higher primes are excluded already by $n \lt 68$, and the pair $55,66$ seems hard to replace. Note that these examples were found by hand, and that analysis of nice subsets of multiples of primes $p_i$ can be done programmatically. I disagree with Jeremy Rouse regarding the level of intractability of this problem, and thank him again for his inspiring example.

**Edit 2017.03.06 GRP**: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out of the good subset, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

Based on gut feel and some trial nomogramming, I suspect the next lowest admissible $n$ will be around $170$ or greater. So far the solutions above are found by hand. The exploration continues. **End Edit 2017.03.06 GRP.**

Gerhard "Who Wants To Go Further?" Paseman, 2017.03.03.