The numbers $e^{H_N-\lfloor H_n\rfloor}$ all fall in the interval $[1,e]$. They do not have a limiting distribution but, in a sense which I will try to make precise, with some smoothing they probably have the uniform distribution in that interval. Otherwise put, the "best" distribution to associate with the given problem assigns the interval $[a,b]$ the probability $\frac{e^b-e^a}{e-1}.$

Let $r(a,b,N)= \frac{A(a,b,N)}{N}.$ It turns out that $$\limsup r(a,b,N) =\frac{e^{1-b}(e^b-e^a)}{e-1} \text{ and }\liminf r(a,b,N)=\frac{e^{-a}(e^b-e^a)}{e-1}.$$ So if $b-a$ is small, then the ratio is close to $e$.

The value of $H_n$ is quite close to $\ln{n}+\gamma$. where $\gamma = 0.577\cdots$ is the Euler Mascheroni constant. As $N$ increases and the values $H_N$ go to infinity at an ever decreasing rate, the count $A(a,b,N)$ stays constant for long stretches during which the ratio $r(a,b,N)$ decreases. For each positive integer $K$ there is a local minimum of $r(a,b,N)$ just before $H_N \gt K+a.$ When that occurs $N \approx e^{K+a-\gamma}.$ Then for a while, every time $N$ increases by $1$ so does $A(a,b,N)$ as well as $r(a,b,N).$ This continues until $H_N \gt K+b,$ which is at $N \approx e^{K+b- \gamma}.$ That is when $r(a,b,N)$ has a local maximum. Finding what those local extreme values are involves summing finite geometric progressions. I did this and compared the results to numerical data ($K=7,8,9$) with which it agrees well.

Notes:

- The same upper and lower limits occur (and are easier) if one replaces $H_n$ by $\int_1^n\frac{1}{x}dx=\ln{n}$.
- Until $\frac{1}{N}$ is smaller than $b-a,$ we might have $H_N \lt K+a \lt K+b \lt H_{N+1}.$ Until it is "much" smaller, things might be less regular. This has very little effect on the convergence.
- The variation where we replace $H_N$ with a random number between $H_N$ and $H_{N+1}$ would be the same.

There are various transformations which take every convergent sequence to another with the same limit but faster convergence. When applied to a divergent but bounded sequence they may take it to another sequence with a limit (and various transformations should all give the same limit). A standard transformation is to replace $\lim r(a,b,N)$ with the limit of an average $$\lim \frac{r(a,b,1)+r(a,b,2)+\cdots + r(a,b,N)}{N}$$ I suspect that *that* might turn out to have limit $\frac{(e^b-e^a)}{e-1}.$ It might not be hard, but I have not checked it carefully.