Can this type of series retain the same value? Let $H$ be a Hilbert space and $\sum_k x_k$  a countable infinite sum in it. Lets say we partition the sequence $(x_k)_k$ in a sequence of blocks of finite length and change the order of summation only in those blocks, like this (for brevity illustrated only for the first two blocks $$(x_1,\ldots,x_k,x_{k+1},\ldots,x_{k+l},\ldots )$$ becomes
$$(x_{\pi(1)},\ldots,x_{\pi(k)},x_{\gamma(k+1)},\ldots,x_{\gamma(k+l)},\ldots ),$$ where $\pi$ and $\gamma$ are permutations.
If we denote the elements of the second sequence with $x'$, does anyone know, what will happen to the series $\sum _k x'_k$ in this case ? Can it stay the same ? Does staying the same requires additional assumptions ?

If both series converge, it doesn't change anything. This can be easily seen by considering partial sums.
Put $k_j$ as the cumulative length of the first $j$ blocks. Then clearly $\sum_{j=1}^{k_n} x_j=\sum_{j=1}^{k_n} x_j'$ for any $n$, so assuming both series converge, we have that 
$$\sum_j x_j=\lim_{n\to \infty}\sum_{j=1}^{k_n} x_j=\lim_{n\to \infty}\sum_{j=1}^{k_n} x'_j=\sum_j x_j'$$
On the other hand, we can change a (conditionally) convergent series into a divergent one using this method – if we take the alternating harmonic sequence $x_n=(-1)^n/n$, sorting large enough blocks so that all the positive elements come before all the negative elements, we can get infinitely many arbitrarily large “jumps”, preventing convergence.
On another note, if the length of permuted blocks is bounded, then I think this kind of thing could not happen (again, by considering partial sums). 
