Is there any double series that cannot exchange the sum?

Let $u_{mn} = 1/(m^2 - n^2)$ if $m \neq n$ and $u_{mn} =0$ if $m = n.$

Note that

$$\frac{\pi^2}{12} = \sum_{n=1}^\infty \sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} \neq \sum_{m=1}^\infty \sum_{n=1, n \neq m}^\infty \frac{1}{m^2 - n^2} = - \frac{\pi^2}{12}.$$

By anti-symmetry, the double sum changes sign with an interchange of indices.

To find the sum, use

$$\begin{align}\sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} &= \lim_{M \to \infty} \frac{1}{2n} \sum_{m=1, m \neq n}^M \left(\frac{1}{m-n} - \frac{1}{m+n} \right) \\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\sum_{k=1}^{M-n} \frac{1}{k} - \sum_{k=1}^{n-1} \frac{1}{k} - \sum_{k=n+1}^{M+n} \frac{1}{k}\right)\\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\frac{1}{n}- \frac{1}{M-n+1} - \ldots - \frac{1}{M+n} \right) \\ &= \frac{1}{2n^2} \end{align}$$

Let $$u_{n,p} = \begin{cases} 1 & \text{ if } n = p+1,\\ -1 & \text{ if } n = p,\\ 0 & \text{ otherwise}. \end{cases}$$ On the one hand, we have: $$\begin{align} \sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p} &= u_{0,0} + \sum_{n=1}^{\infty}(u_{n,n} + u_{n,n-1}) &\text{(all other terms are 0)}\\ & = -1 + \sum_{n=1}^{\infty}(-1 + 1) &\text{(evaluate the terms)}\\ & = -1 \end{align}$$ On the other hand, we have: $$\begin{align} \sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p}& = \sum_{p=0}^{\infty}(u_{p,p} + u_{p+1,p}) &\text{(all other terms are 0)}\\ &= \sum_{p=0}^{\infty}(-1 + 1) &\text{(evaluate the terms)}\\ &= 0 \end{align}$$