I am trying to solve this problem.
If $$a_{m,n} = \frac{m-n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}, (m, n > 0)$$ $$a_{m,0}=2^{-m}, a_{0, n} = -2^{-n}, a_{0, 0} =0,$$ Show that $\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right) = -1, \sum_{n=0}^{\infty}\left(\sum_{m=0}^{\infty}a_{m,n}\right) = 1$.
Here is my attempt for $\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)$.
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)= \sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}.$$ Then, we have
\begin{align*} \sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right) &= \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}\left[\frac{m-n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]+\frac{1}{2^m}\right) \\ &= \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}\left[\frac{m}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]-\sum_{n=1}^{\infty}\left[\frac{n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]+\frac{1}{2^m}\right) \\ &= \sum_{m=1}^{\infty}\left(1-\frac{1}{2^m}-1+\frac{1}{2^m} \right) \\ &= 0. \end{align*}
See below about how I get $0$.
\begin{align*} \sum_{n=1}^{\infty}\frac{m}{2^{m+n}}\frac{(m+n-1)!}{m!n!} &= \frac{1}{2^m} \sum_{n=1}^{\infty} \frac{(m+n-1)!}{(m-1)!n!}\left(\frac{1}{2}\right)^n \\ &= \frac{1}{2^m} \sum_{n=1}^{\infty}{m+n-1 \choose n}\left(\frac{1}{2}\right)^n \\ &= \frac{1}{2^m} \sum_{n=1}^{\infty}(-1)^n{-m \choose n}\left(\frac{1}{2}\right)^n \\ &= \frac{1}{2^m}\left[(1-\frac{1}{2})^{-m}-1\right] (\text{binomial expansion})\\ &= \frac{1}{2^m}\left[2^m-1\right] \\ &= 1-\frac{1}{2^m}. \end{align*}
Similarly, \begin{align*} \sum_{n=1}^{\infty}\frac{n}{2^{m+n}}\frac{(m+n-1)!}{m!n!} &= \frac{1}{2^m}\sum_{n=1}^{\infty}\frac{1}{2^{n}}\frac{(m+n-1)!}{m!(n-1)!} \\ &= \frac{1}{2^m}\sum_{n=1}^{\infty}{m+n-1 \choose n-1}\frac{1}{2^{n}} \\ &= \frac{1}{2^m}\sum_{n=1}^{\infty}(-1)^{n-1}{-(m+1) \choose n-1}\frac{1}{2^{n-1}}\frac{1}{2} \\ &=\frac{1}{2^m}\frac{1}{2}\left(1-\frac{1}{2}\right)^{-(m+1)} \\ &= \frac{1}{2^{m+1}}2^{m+1}\\ &=1. \end{align*} Thus, we have
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)= \sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n} = 0-1.$$
I just answer my own question.
Let $b_{m,n}=\frac{(m+n-1)}{2^{m+n}m!n!}$ so $a_{m,n}=(m-n)b_{m,n}$. Yhe sums $(n=1,\infty),(m=1,\infty)$ and switched, over $mb_{m,n}$ and $nb_{m,n}$ are divergent. However the sums over $(n-m)b_{n,m}$ are both zero.
The final pieces are $\sum_{m=1}^\infty \frac{1}{2^m}=1$ and $\sum_{n=1}^\infty -\frac{1}{2^n}=-1$, which cancel.