Let $\displaystyle \{z_1,z_2, \ldots, z_n\}$ be $n$ complex numbers such that: $\displaystyle \sum\limits_{k=1}^n|z_k| = 1$
Then we have to show that, there is a subset $S$ of $\{1,2,\ldots,n\}$, such that, $\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{\pi}$.
Owing to the symmetry of the expression its easy to see: $$\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{4}$$ Since, $1 = \sum\limits_{k=1}^{n} |z_k| \le \sum\limits_{\mathfrak{Re}(z) \ge 0} |\mathfrak{Re} (z_k)|+\sum\limits_{\mathfrak{Re}(z) < 0} |\mathfrak{Re} (z_k)|+\sum\limits_{\mathfrak{Im}(z) \ge 0} |\mathfrak{Im} (z_k)|+\sum\limits_{\mathfrak{Im}(z) < 0} |\mathfrak{Im} (z_k)|$, one of the four summations on the RHS must exceed the others, say wlog it is $\mathfrak{Re}(z) \ge 0$,
then, $\left|\sum\limits_{\mathfrak{Re}(z_k) \ge 0} z_k \right| \ge \left|\sum\limits_{\mathfrak{Re}(z_k) \ge 0} \mathfrak{Re}(z_k) \right| \ge \frac{1}{4}$.
(1) But how do we reach the lower bound $\displaystyle \frac{1}{\pi}$ ?
If we remove the homogeneity condition, we are required to prove, $\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{\pi}\sum\limits_{k=1}^n|z_k|$ for a subset $S$ of $\{1,2,\ldots,n\}$.
(2) How do we determine if $1/\pi$ is the best constant here ?
