If the sizes of the blocks are bounded, then the permuted summation also converges to the same limit.
More generally, let $\sigma : \Bbb{N}_0 \to \Bbb{N}_0$ be a permutation such that $M = \sup_{n\in\Bbb{N}_0}|\sigma(n) - n|$ is finite. You can easily verify that this generalizes the former special case. Let $x'_n = x_{\sigma(n)}$ be the permuted sequence and $A_n = \{0, \cdots, n\}$. Now, from the assertion we find that
- $\sigma(A_n) \subset A_{n+M}$ and likewise $\sigma^{-1}(A_n) \subset A_{n+M}$, and
- $A_{n+M} \setminus \sigma(A_n) \subset A_{n+M}\setminus A_{n-M}$.
The first claim is trivial. To prove the second one, just observe that for $n > M$,
$ A_{n+M}\setminus \sigma(A_n) \subset A_{n+M}\setminus \sigma(\sigma^{-1}(A_{n-M})) = \{n-M+1, \cdots, n+M\}. $
Now we are ready. Let $(s_n)$ and $(s'_n)$ denote the partial sum of $(x_n)$ and $(x'_n)$, respectively. Then
$\begin{align*}|s_n - s'_n| &\leq |s_{n+M} - s_n| + |s_{n+M} - s'_n| \\ &\leq \sum_{k=n+1}^{n+M} |x_k| + \sum_{k \in A_{n+M}\setminus \sigma(A_n)}|x_k| \\ &\leq \sum_{k=n+1}^{n+M} |x_k| + \sum_{k = n-M+1}^{n+M} |x_k|. \end{align*}$
Now there are only $3M$ terms of $(x_k)$, each of which converges to 0 as $n \to \infty$. Thus by squeezing lemma, $|s_n - s'_n| \to 0$. This shows that
$ \sum_n x'_n = \lim_n s'_n = \lim_n s_n = \sum_n x_n.$
If $M$ is unbounded, you can easily devise a counter-example. For example, if we take the sizes of the blocks as $2, 4, 6, \cdots$ to the conditionally convergent series
$ 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \cdots = 0 $
and permute the order of the sequence in each block as
$ 1 - 1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{3} + \cdots, $
it oscillates between 0 and 1.