Given a Banach space $X$, and an operator $T \in X^*$, when does it hold that $$\sum_{i=1}^\infty |T(x_n)| < \infty$$ for every sequence $\{x_n\}$ in $X$ such that $\sum_1^\infty x_n$ converges? As @Conrad points out, there is the trivial counter-example of the identity map on $X = \mathbb C$.
Note that I am not requiring absolute convergence of the $\sum x_i$. The problem would be trivial otherwise.
Context: I am a novice at Banach space theory, and this problem came up when I was studying another problem. In particular, fix a measurable space $(X,\mathcal M)$, and let $S$ be the Banach space of all bounded $\mathcal M$-measurable functions $f : X \to \mathbb C$ equipped with the uniform norm. What conditions must $T \in S^*$ obey for there to be a complex measure $\mu$ such that $$Tf = \int f\,d\mu \quad \forall f \in S?$$ If $T$ were to have the magical property I described above, then this would be enough, because then one could define $\mu(E) = T\chi_E$. However, this only appears to work when $T$ has this property. To see this, by continuity, if $\{E_n\} \subset \mathcal M$ is a countable disjoint partition of a set $E \in \mathcal M$, then since $\chi_E = \sum \chi_{E_n}$ and the convergence is uniform (but notably not absolute w.r.t. the uniform norm), continuity dictates that $$\mu(E) = T\chi_E = \sum_{i=1}^\infty T\chi_{E_i} = \sum_{i=1}^\infty \mu(E_i).$$ But the convergence is absolute only if $$\sum_{i=1}^\infty |T\chi_{E_i}| < \infty,$$ and this happens when $T$ satisfies this magical property. In particular this means that $\mu$ is a well-defined measure, and it is not too hard to work out that $Tf = \int f \,d\mu$.
Pick any $x$ such that $Tx \neq 0$. Then take the sequence $x,-x,{1 \over 2}x , -{1 \over 2}x,..., {1 \over n}x, - {1 \over n}x,...$. Clearly the sum converges, but $\sum_k |T(x_k)| = \infty$.