I thank Gerry Myerson. His post put me on watch for papers coauthored by Silberger, of which a recent one (http://arxiv.org/abs/1702.01316) refers to a paper of Erdos and Niven (Some properties of partial sums of the harmonic series) from 1946 which answers the title question affirmatively. Thanks also to the Silbergers and Erdos and Niven.

I thank Michael Stoll for his interest in the problem and for clarifying some statements I made in the post.

I also thank Włodzimierz Holsztyński for inspiring the following proof sketch, which solves the version in the edit. (I begin the sketch now, and will finish it and provide another in a subsequent edit.) The motivation is to show for integers $0 \lt k \lt n$ and $0 \lt a \lt b$ that $H(ka,kb) \neq H(na,nb)$. Using $a=0$ and $b=1$ for guidance, I show $H(ka,kb) \lt H(na,nb)$ for $b=a+1$, which will imply the motivated result.

I start by splitting each of the $k$ summands of $H(ka,ka+k)$ into $n$ equal parts, and similarly the $n$ summands of $H(na,na+n)$ each into $k$ equal parts. That suggests comparing two sums each of $nk$ terms. Except that I am using $a$ instead of $b$, and that I want $j=nk$, the sum comparison looks exactly like the comparison of the question edit above. Of course, if I prove the conjecture in the edit, I get the result I want.

I change notation now replacing $a$ and $a+1$ with $b$ and $b+1$, and subtract the two sums term by term. It is now enough for me to show that the following holds for $0 \lt j \leq nk$:

$$\sum^j_{i=1} \frac{n \lceil i/n \rceil - k \lceil i/k \rceil}{k(nb + \lceil i/k \rceil)n(kb + \lceil i/n \rceil)} \gt 0$$

This is a straightforward induction on $j$ when $t_j = n \lceil j/n \rceil - k \lceil j/k \rceil$ is positive.
Since $t_1$ is positive, the partial sum can be shown positive also when $t_j$ is zero and the induction holds. It remains to show that the sum is still positive for those $j$ for which $t_j$ is negative. This will be provided in the edit to come.

**Edit 2017.02.07 GRP**:
To handle the case when $t_j$ is negative, I prove something even stronger. I restrict the indices $i$ and $j$ to range between $mn+1$ and $mn+n$ for an integer $m$ with $0 \leq m \lt k$, and I show that this portion (the sum for $i=mn+1$ to $j$) is also positive. I look for the largest $i$ in this range with $t_i \geq 0$ and note that it is not far from $mn+n$. If $r =(mn+n) \bmod k$ with $0 \leq r \lt k$ then this largest $i$ is $mn+n-r$. We now have $(n-r)$ positive terms of weight $r$ or more against $j-(n-r) \leq r$ negative terms of weight at most $(k-r)$ (I am suppressing mention of the denominators and the exact numerators of the summands), and since $n \gt k$, this means the positive terms outweigh the negative terms. Thus the subsubsums are also positive, so the subsums are positive, giving the desired conclusion and the end of this sketch.

I told you that story to tell you this one. The just completed sketch shows promise in giving an elementary proof of the title problem in a similar way. The hitch is that an additional term shows in the numerator. I will start this sketch and hope to finish it in a subsequent edit.

Recall that it suffices to show that subsums sharing no terms are disjoint. Now choose positive integers $a$ and $x$, and then choose $b \geq a+x$. There will be a unique integer $y \gt x$ such that $H(b,b+y-1) \lt H(a,a+x) \leq H(b,b+y)$. I want to show $H(a,a+x) \lt H(b,b+y)$ again by splitting the $1/(a+i)$ summands into $y$ equal pieces, and similarly each $1/(b+i)$ in $x$ equal pieces. I then line up the sums of the $xy$ many pieces and hope to prove positivity of the difference by proving positivity of the subsums as in the previous sketch.

When I do this, I get the following as a summand, where I use $d=ya-xb$:
$$ \frac{d+ y\lceil i/y \rceil -x\lceil i/x \rceil}{y(a+ \lceil i/y \rceil)x(b + \lceil i/x \rceil)} .$$
Now if $d \geq 0$, the sketch above goes through and we can celebrate. If $d$ is negative and small, I think a variant of the above sketch will work, right down to the subsubsum portion. However I do not have that to put in this edit.

If $y/b$ is small (and indeed I can use Bertrand to make it less than 1 and other results in prime gaps to make it less than 1/5) then I can show $d$ has small size (perhaps less than $(y-x)/2$), and I may be able to tweak the sketch to provide a new and weaker proof of the Erdos Niven result. I would like to make it even more elementary, so with this edit I invite others to play with it.
**End Edit 2017.02.07 GRP**

Gerhard "Serializing Answers Seems Somewhat Easier" Paseman, 2017.02.07.