It helps to visualize the terms in a square of products $1/(ij)$ with $i$ and $j$ running for $1$ to $n$. The sum over the first term contains all products with $i\ne j$ exactly once, whereas the sum over the second term roughly corresponds to the upper left half of the square, but with the left-most column, which adds up to $H_n$, excluded. Thus we have
$ \def\sub#1{{\scriptstyle{i\ne j}\atop{\scriptstyle i,j\le #1}}} \sum_{k=1}^{n-1}\frac{2}{k+1}H_k=\sum_{\sub n}\frac1{ij}$
and
$-\sum_{k=1}^{n-1}\frac{1}{1+n-k}=H_n-\sum_{i+j\le n+1}\frac1{ij}\;.$
Substituting this into your equation, multiplying through by $n-1$ and simplifying leads to
$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=\frac2{n+1}H_n\;,$
$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=2\sum_i\frac1{n+1}\frac1i\;,$
$\sum_{i+j\le n+1}\frac1{ij}=\sum_{\sub{n+1}}\frac1{ij}\;.$
This we can prove by induction: The equation is satisfied for $n=0$, and going from $n$ to $n+1$ adds
$\sum_{i+j=n+1}\frac1{ij}=\sum_{i=1}^n\frac1i\frac1{n+1-i}$
to the left-hand side and also
$2\frac1{n+1}\sum_{i=1}^n\frac1i=\frac1{n+1}\sum_{i=1}^n\left(\frac1i+\frac1{n+1-i}\right)=\sum_{i=1}^n\frac1i\frac1{n+1-i}$
to the right-hand side.