I'm interested in the following combinatorial problem: What is a necessary and sufficent condition on a permutation $\sigma : \mathbb{N} \rightarrow \mathbb{N}$, so that there exist a summable sequence of real numbers $a_n$ for which $\Sigma a_{\sigma(n)}$ converges but $\Sigma a_{\sigma(n)} \neq \Sigma a_n$.

Call an interval of $\mathbb{N}$ a set of consecutive integers. A necessary condition, which follows from applying the Cauchy criterion in both ways is the following: There does not exist a sequence (indexed by $i$) of sets of intervals $\{A^i_1,A^i_2....,A^i_{m_i}\}$, and $\{B^i_1,B^i_2....,B^i_{n_i}\}$ s.t $\{n_i\}$ and $\{m_i\}$ are bounded, $\sigma(\cup_{r=1}^{m_i} A^i_r)=\cup_{r=1}^{n_i} B^i_r$ for all $i$, and the sequence $\{\cup_{r=1}^{m_i} A^i_r\}$ forms an increasig filtration of $\mathbb{N}$.

I'm unsure of wheter the above condition is sufficent however.

Edit: Also, I'm aware of this question which answers when a permutation sends all convergent series to convergent series with the same sum. However, it doesn't quite answer this question as the complement to that set includes many permutations which send convergent series to non convergent series, but which don't send any convergent series to a convergent series with a different value. From my testing, it seems that being able to alter the value of a series while still keeping it convergent is a far more delicate issue than just being able to make a series not converge.