jmc

the sum of the first $m$ positive odd integers is $$1+3+5+\cdots +(2m-1)=m^2,$$and the sum of the first $n$ positive even integers is $$2+4+6+\cdots +2n=n^2+n,$$so $m^2-(n^2+n)=212.$ Then $$4m^2-(4n^2+4n)=848,$$so $4m^2-(2n+1{)}^2=847.$ By difference of squares, $$(2m+2n+1)(2m-2n-1)=847.$$We list the ways of writing 847 as the product of two positive integers, and the corresponding values of $m$ and $n.$ (Note that $2m+n+1$ must be the larger factor.)

$$\begin{array}{cccc}2m+2n+1& 2m-2n-1& m& n\\ 847& 1& 212& 211\\ 121& 7& 32& 28\\ 77& 11& 22& 16\end{array}$$Thus, the sum of the possible values of $n$ is $211+28+16=\boxed{255}.$