It follows from Bertrand's postulate that $mH_m - nH_n$ is an integer only in the case $n=m$. Suppose $n<m$, and assume that $m$ is reasonably large below (say $\ge 20$).

Let $p$ denote the largest prime below $m$, so that by Bertrand's postulate $p>m/2$. The only term with $p$ in the denominator in $mH_m$ is $m/p$ which is not an integer (since $m/2 <p <m$). So if there is no term with $p$ in the denominator in $nH_n$, then $mH_m -nH_n$ cannot be an integer. This takes care of the case when $p>n$.

If $p\le n$, then $nH_n$ has the unique term $n/p$ with $p$ in the denominator. But the only way $m/p - n/p$ can be an integer is if $n=(m-p)$ and we also have $p\le n$, but this is impossible since $p >m/2$.

It follows from Bertrand's postulate that $mH_m - nH_n$ is an integer only in the case $n=m$. Suppose $n<m$, and assume that $m$ is reasonably large below (say $\ge 20$).

Let $p$ denote the largest prime below $m$, so that by Bertrand's postulate $p>m/2$. The only term with $p$ in the denominator in $mH_m$ is $m/p$ which is not an integer (since $m/2 <p <m$). So if there is no term with $p$ in the denominator in $nH_n$, then $mH_m -nH_n$ cannot be an integer. This takes care of the case when $p>n$.

If $p\le n$, then $nH_n$ has the unique term $n/p$ with $p$ in the denominator. But the only way $m/p - n/p$ can be an integer is if $n=(m-kp)$ (for some integer $k\ge 1$) and we also have $p\le n$, but this is impossible since $p >m/2$.