Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.  
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_m \sum_{k=1}^m - \sum_{k=1}^{m-1} (a_{k+1}-a_k)\sum_{i=1}^k b_i \big)$, but it leads to nowhere.

This solution is similar to an inductive proof with respect to $m$.
Let the sum of interest be
$$s_{n,m}=\frac{1}{\binom{n}{m}} \sum _{k=1}^m \frac{1}{k} \binom{n-k}{n-m}$$
For $m=1$ we find
$$s_{n,1}=\frac{1}{n}$$
The difference of $s$ with respect to $m$ turns out to be surprisingly simple:
$$s_{n,{m+1}}-s_{n,m}=\frac{1}{n-m}\tag{1}$$
so that
$$s_{n,2}=\frac{1}{n}+\frac{1}{n-1}$$
$$s_{n,3}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}$$
$$...$$
$$s_{n,m}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+...+\frac{1}{n-m+1}$$
The last expression is easily identified as the difference between two harmonic numbers. Hence
$$s_{n,m}=H_{n}-H_{n-m}$$
as requested.
And now we prove the recurence relation (1) using the definition
$$\binom{n}{m}=\frac{n!}{m!\;(n-m)!}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}
s_{n,{m+1}}-s_{n,m} 
&=& \frac{1}{\binom{n}{m+1}} \sum_{k=1}^{m+1} \frac{1}{k}\binom{n-k}{n-m-1} - \frac{1}{\binom{n}{m}} \sum_{k=1}^{m} \frac{1}{k}\binom{n-k}{n-m}\\
&=& \frac{1}{\binom{n}{m}}\left( \frac{m+1}{n-m}\left( \sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m-1} +\frac{1}{m+1}\right) -\sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m}\right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{k}\binom{n-k}{n-m} \left( \frac{m+1}{m+1-k} -1 \right) \right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \frac{n-m}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \binom{n-k}{m-k+1} \right) =\frac{1}{n-m} \\
\end{eqnarray*}$
In the last step we have used the identity
$\begin{eqnarray*}
\binom{n}{m}
&=&  \sum_{k=1}^m \binom{n-k}{m-k+1} +1 = \sum_{k=1}^{m+1} \binom{n-k}{m-k+1} \\
\end{eqnarray*}$
which in turn results from iterating the basic recursion of the binomial coefficients
$\begin{eqnarray*}
\binom{n}{m}
&=& \binom{n-1}{m}+\binom{n-1}{m-1} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+\binom{n-2}{m-2} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+...+\binom{n-m}{1} +\binom{n-m-1}{0}\tag{2}\\
\end{eqnarray*}$
This completes the proof of (1) and hence the solution.
Comments
1) I feel that a much shorter proof should be possible starting from (2), but I didn't find it.   
2) Interesting related formula (Derivative of binomial coefficients)
$$\frac{d}{dn}\ln\binom{n}{k} = H_{n} - H_{n-k}$$
