Fix $k\in\mathbb{N}.$ Consider a sequence of non-intersecting balls $\left(B(x_n,r_n)\right)_{n\in\mathbb{N}}$ in $\mathbb{R^k}:\quad p\neq q \implies B(x_p,r_p) \cap B(x_q,r_q) = \emptyset.$
Proposition: If $\exists\ (a_n)_{n\in\mathbb{N}} $ with $a_n\in B(x_n,r_n)\ \forall\ n\in\mathbb{N},\ $ such that $\displaystyle\sum_{n=1}^{\infty} a_n\ $ converges, then $\displaystyle\sum_{n=1}^{\infty} b_n\ $ converges for every $(b_n)_{n\in\mathbb{N}}\ $ with $b_n\in B(x_n,r_n)\ \forall n\in\mathbb{N}. $
Obviously the origin cannot be in any of the balls, for then $\displaystyle\sum_{n=1}^{\infty} a_n\ $ would not converge.
If we use large balls, then this limits space, making a counter-example less likely. Small balls going in a spiral inwards toward the origin might give a counter-example, but I haven't done this successfully yet.
For the simplicity sake, we assume that all balls $B(x_n,r_n)$, $n\in\mathbb{N}$ are open.
There is a counterexample for Proposition already for $k=2$ constructed as follows. Put $A=\frac{\pi^2}6$. It is well known that $A=\sum_{i=1}^\infty \frac 1{i^2}$, so we can partition the segment $[0,A]\times \{0\}\subset \mathbb R^2$ from its right endpoint into consecutive segments $I_1,I_2,\dots,$ such that the length of $I_i$ is $\frac 1{i^2}$ for each natural $i$. Next, on each segment $I_i$ we build a vertical stack of $i$ balls of diameter $\frac 1{i^2}$ each. Now let $\{a_{2n-1}:n\in\mathbb N\}$ be any enumeration of the centers of the built balls, such that the diameter of each next ball is nonincreasing. Moreover, for each natural $n$ put $a_{2n}=-a_{2n-1}$. Then the set $\{a_{2n}:n\in\mathbb N\}$ is the set of centers of balls symmetric to the already constructed balls with respect to the origin $(0,0)$ of the plane $\mathbb R^2$. It is easy to see that $\sum_{n=1}^\infty a_n=(0,0)$. Now for each natural $n$ put $b_{2n}=a_{2n}$ and $b_{2n-1}=a_{2n}+\left(\frac{1}{3i^2},0\right)$, where $I_i$ is the segment over which is placed the ball with the center $a_{2n}$. Then $$\sum_{n=1}^\infty b_n =\left(\sum_{i=1}^\infty \frac{i}{3i^2}, 0\right)=(\infty,0).$$
On the other hand, Proposition holds for $k=1$. Indeed, in this case the sequence $(a_n)_{n\in\mathbb{N}}$ converges to zero, so there exists $M\in\mathbb R$ such that $|a_n|\le M$ for each natural $n$. Put $A=\{n\in\mathbb N: |x_n|>M\}$. Since the segments $\left(B(x_n,r_n)\right)_{n\in\mathbb{N}}$ are pairwise disjoint and intersect the segment $[-M, M]$, is it easy to see that there is at most one natural $\ell$ and at most one natural $r$ such that $x_\ell<M$ and $x_r>M$, that is $|A|\le 2$. Since $r_n\le 2M$ for each $n\in\mathbb N\setminus A$, the union of the pairwise disjoint segments $\left(B(x_n,r_n)\right)_{n\in\mathbb{N}}$ is bounded, so the sum $\sum_{n=1}^\infty 2r_n$ of their diameters is bounded too, that is the series $\sum_{n=1}^\infty r_n$ converges, and so does the series $\sum_{n=1}^\infty b_n$, because the series $\sum_{n=1}^\infty a_n$ converges and $|b_n-a_n|\le 2r_n$ for each natural $n$.