Suppose that $n=2k+1$ and $m=2j+1$; then
$\begin{align*} \frac1{n^2-m^2}&=\frac1{2n}\left(\frac1{n-m}+\frac1{n+m}\right)\\ &=\frac1{2n}\left(\frac1{2(k-j)}+\frac1{2(k+j+1)}\right)\\ &=\frac1{4n}\left(\frac1{k-j}+\frac1{k+j+1}\right)\;. \end{align*}$
Now split the sum into the terms with $m and the terms with $m>n$ and write it as
$\frac1{4n}\left(\sum_{j=0}^{k-1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j>k}\left(\frac1{k-j}+\frac1{k+j+1}\right)\right)\;.$
Now $\sum_{j=0}^{k-1}\frac1{k-j}=\sum_{i=1}^k\frac1i\;,$ and $\sum_{j=k+1}^{2k}\frac1{k-j}=-\sum_{i=1}^k\frac1i\;,$ so these terms cancel out, and we’re left with
$\begin{align*} &\frac1{4n}\left(\sum_{j=0}^{k-1}\frac1{k+j+1}+\sum_{j\ge 2k+1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j=k+1}^{2k}\frac1{k+j+1}\right)\\ &\qquad=\frac1{4n}\left(\left(\sum_{j\ge 0}\frac1{k+j+1}-\frac1{2k+1}\right)+\sum_{j\ge 2k+1}\frac1{k-j}\right)\\ &\qquad=\frac1{4n}\left(\sum_{i\ge k+1}\frac1i-\frac1n-\sum_{i\ge k+1}\frac1i\right)\\ &\qquad=-\frac1{4n^2}\;. \end{align*}$