Suppose $(z_n)_{n\in\mathbb{N}}\subset \mathbb{C}$ such that $\displaystyle\sum_{n=1}^{\infty} z_n$ converges.
Then, for example,
$$ \sum_{k=1}^{\infty}\left( \sum_{\frac{1}{k+1} < \left\lvert z_n \right\rvert < \frac{1}{k} }^{} z_n \right) $$
does not always converge, for example consider:
$$z_1=1,\ z_2=z_3=\ldots=z_{10^{10}+1} = -\frac{1}{10^{10}},\ z_{10^{10}+2} = \frac{1}{3},\ z_{10^{10}+3}=z_{10^{10}+4}=\ldots = z_{10^{10}+2 + 10^{10^{10}}}= -\frac{1}{10^{10^{10}}}, \ldots $$
But this prompts the question that, given the assumptions at the start of the question:
Does there exist a sequence $(\phi_n)_{n\in\mathbb{N}}$ of nested Jordan curves (i.e. closed simple curves) in $\mathbb{C};\ $ nested in the sense that $\phi_{k+1}$ is contained in the interior, $I_k,$ of $\phi_k;$ and the curves tend towards zero, i.e. all the curves contain the point $0+0i$ in the interior, and $\text{diam}(\phi_n)\to 0;$ such that
$$ \sum_{k=1}^{\infty}\left( \sum_{ z_n \in I_k\setminus I_{k+1} } z_n \right) $$
converges?
The answer is yes for the specific sequence above. In fact, since all members of the above sequence are in $\mathbb{R},$ you could say that the version of this question in the realm of $\mathbb{R}$ is the following:
Suppose $(x_n)_{n\in\mathbb{N}}\subset \mathbb{R}$ such that $\displaystyle\sum_{n=1}^{\infty} x_n$ converges.
Do there exist sequences $(a_n), (b_n)\subset\mathbb{R}$ such that $\ a_n < a_{n+1} < 0 < b_{n+1}<b_n\ \forall\ n\in\mathbb{N},\quad a_n\to 0^-,\ b_n\to 0^+,\ $ and
$$ \sum_{k=1}^{\infty}\left( \sum_{ x_n \in (a_k,a_{k+1}] \cup (b_{k+1},b_{k}] } x_n \right) $$
converges?
Clearly, if the answer to the first question is yes, then the answer to the second question is also yes.
But I'm not even sure if the answer is yes for the $\mathbb{R}$ case, although maybe you can prove it via greedy algorithm in some way.
Let $(I_k)_{k\in\mathbb N}$ be the sequence of open subsets of $\mathbb C$. Assume without loss of generality that $\overline{I_{k+1}}\subset I_k$, which you can do because the diameters of these sets go to $0$. So $(I_k\setminus \overline{I_{k+1}})\cap\mathbb R$ is an open subset of $\mathbb R$ and it includes both positive and negative numbers since $0\in I_{k+1}$.
Observe that the sums $$ \sum_{ z_n \in I_k\setminus I_{k+1} } z_n $$ don't depend on the ordering of the sequence $(z_n)_n$. So we can choose a sequence $(x_n)_n$ satisfying $$\sum_{ x_n \in I_k\setminus I_{k+1} } x_n \approx (-1)^k$$ by picking enough copies of any element of $(I_k\setminus \overline{I_{k+1}})\cap\mathbb R_{>0}$ or $(I_k\setminus \overline{I_{k+1}})\cap\mathbb R_{<0}$ to make the sum exceed $1$ in absolute value and then order them in a convenient way. The condition on the diameters of the $I_k$ implies that $\lim_{n\to\infty}x_n=0$ and then letting $z_n=x_{\sigma(n)}$ where $\sigma:\mathbb N\to\mathbb N$ is a bijection chosen so that $z_{n+1}$ is the largest absolute value element left with sign opposite to $\sum_{h=1}^n z_h$. We can always do this because $\sum_{n:x_n>0}x_n=\infty$ and $\sum_{n:x_n<0}x_n=-\infty$. This guarantees that $$\left|\sum_{h=1}^n z_h\right|\leq |z_n|\to 0.$$