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 $mn$ 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
$$\begin{align*}
&\frac1{4n}\left(\sum_{j=0}^{k-1}\frac1{k+j+1}+\sum_{j\ge 2k+1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j=k+1}^{2k}\frac1{k+j+1}\right)\\
&\qquad=\frac1{4n}\left(\left(\sum_{j\ge 0}\frac1{k+j+1}-\frac1{2k+1}\right)+\sum_{j\ge 2k+1}\frac1{k-j}\right)\\
&\qquad=\frac1{4n}\left(\sum_{i\ge k+1}\frac1i-\frac1n-\sum_{i\ge k+1}\frac1i\right)\\
&\qquad=-\frac1{4n^2}\;.
\end{align*}$$