Maximum value $c$ s.t. $\exists$ a subset $S$ of $\{z_1,z_2,\ldots,z_n\}$ s.t. $\left|\sum_{z\in S}z\right|\geq c$ ($\sum_{i=1}^{n}|z_i|=1$). The original question is unclear so I completely rephrased the question and provided full context:
This is the original question (from CMO 1986):

Let $z_1,z_2,\cdots ,z_n$ be complex numbers satisfying $$|z_1|+|z_2|+\cdots +|z_n|=1.$$ Prove that there exists a subset $S$ of $\{z_1,z_2,\ldots,z_n\}$ such that $$\left|\sum_{z\in S}z\right|\geq\frac16.$$

But obviously the lower bound $\frac16$ is far less than optimal. For example, by simply using pigeonhole principle and the inequality $|z|\leq\Re(z)+\Im(z)$ it is easy to show that for every set of complex numbers $\{z_1,z_2,\ldots,z_n\}$ there always exists $S$ such that $\left|\sum_{z\in S}z\right|\geq\frac14$. But I wonder how to obtain the optimal value, i.e.:

Let $z_1,z_2,\cdots ,z_n$ be complex numbers satisfying $$|z_1|+|z_2|+\cdots +|z_n|=1.$$ Find the maximum value $c$ such that for every set of complex numbers $\{z_1,z_2,\ldots,z_n\}$ satisfying the condition above, there always exists a subset $S$ of $\{z_1,z_2,\ldots,z_n\}$ such that $$\left|\sum_{z\in S}z\right|\geq c.$$

I read from the book 101 Algebra Problems from the Training of the USA IMO Team (p85) that "Using advanced mathematics, the lower bound can be
further improved to $\frac1{\pi}$." But how exactly do I obtain that result?
 A: A bit of intuition to start: Our basic idea here will to be to think of our complex numbers as vectors in $\mathbb{R}^2$ and work with projections instead of lengths, mainly because projections work so nicely with respect to sums: The projection of the sum is the sum of the projections.  So instead of trying to create a sum with large norm, we'll pick a direction and try to create a sum with large projection/component along that vector.  
Suppose that I gave you a direction $v$ in advance, and all you wanted to do was maximize the component of your sum along $v$.  Then choosing your subset would be easy: You include the vectors that point "towards" $v$ (have positive inner product), and exclude the ones that point "away" from $v$ (have negative inner product).  
But since I'm the one picking $v$, you may be completely out of luck -- maybe I'll give you a $v$ that points away from all of your vectors, and the best component you could get is $0$.  We want to pick the "right" $v$ in some sense -- one that gives us a large projection.  But it's hard to do that without knowing the vectors in advance.  And furthermore, Han de Bruijn's answer suggests that the extreme case comes when everything's symmetrical and we could just pick any direction whatsoever at random.  
What this suggests is that we should pick a random direction, or to put it differently, show that the average $v$ gives us a large projection.  Now for the actual calculation ...

Let $v$ be an arbitrary unit vector.  As suggested in the intuition above, let $S_v$ denote those indices $i$ for which $z_i$ has a positive inner product with $v$, and define
$$x_v := \sum_{i \in S_v} z_i.$$
We can bound $|x_v|$ from below by its component along $v$:
$$|x_v| \geq \langle x_v, v\rangle = \sum_{i \in S_v} \langle z_i, v \rangle = \sum_{i \in S_v} |z_i| \cos(\theta_{i,v}),$$
where $\theta_{i,v}$ is the angle between $z_i$ and $v$.  We can rewrite this as 
$$|x_v| \geq \sum_{i=1}^n |z_i| \max\{0, \cos(\theta_{i,v})\},$$
  since by definition the extra terms we're including are all just $0$.  
Now suppose I were to pick a $v$ uniformly at random from the unit circle.  Then the angle $\theta_{i,v}$ would be uniform on $[0,2 \pi)$, and the expected value of the $i^{th}$ term on the right hand side would be 
$$\frac{1}{2 \pi} \int_0^{2\pi}  |z_i| \max\{0, \cos \theta\} \, d\theta = \frac{1}{\pi} |z_i|.$$
Adding up over all $i$, we have 
$$E(|x_v|) \geq \frac{1}{\pi} \sum_{i=1}^n |z_i| = \frac{1}{\pi}$$
Here $E(|x_v|)$ denotes the expected value (or average) of $|x_v|$ taking over a randomly chosen $v$.  
The key point here (what's sometimes termed "Erdős magic" after Paul Erdős): If we have a collection of vectors where the average length is $\frac{1}{\pi}$, that means there must be a vector in that collection whose length is at least $\frac{1}{\pi}$, so we win.  That $\frac{1}{\pi}$ is the best possible can be shown using the point set from Han de Bruijn's answer.

This problem is a sort of old chestnut, and the argument here is not my own.  But I don't actually know the original source for it.  I'd love it if anyone who knows a bit more of its history could comment.  
