Prove that $max_{J\subset \{ 1,2...n \} } \left| \sum_{j \in J} z_j\right| = max_{t\in[0,2\pi]} \sum_{j=1}^{n} Re^{+}(e^{it}z_j).$ Consider a finite subset of $\mathbb{C}$ $A=\{z_1,z_2,...,z_n\}$ and the function $Re^{+}(z) = Re (z) $ if $ Re(z)>0$ and $0$ if $Re(z)\leq 0$. 
I have to prove that:
$max_{J\subset \{ 1,2...n \} } \left| \sum_{j \in J} z_j\right| = max_{t\in[0,2\pi]} \sum_{j=1}^{n} Re^{+}(e^{it}z_j).$
Geometrically it means that if you have $n$ points in the plane, the maximum of the sum of the positive $x$ components of the points when you turn them with an angle $t$ is equal to the sum of the distances from the center to the points certain subset of A.

One direction: $\max_{J\subset\{1,\dots,n\}} |\sum_{j\in J} z_j|\le\max_{t\in[0,2\pi]} \sum_{j=1}^n \Re^+(e^{it}z_j)$:
Choose $J$ so that the max on the LHS is attained. Let $t=-\arg\sum_{j\in J} z_j$.
Then $|\sum_{j\in J} z_j|=\sum_{j\in J} \Re(e^{it}z_j)$. I claim that for each $k\in J$, $\Re(e^{it}z_k)\ge0$ and for each $k\notin J$, $\Re(e^{it}z_k)\le0$. To see this, let $k\in J$. Then by the choice of $J$, we have
$$\sum_{j\in J} \Re(e^{it}z_j)=|\sum_{j\in J} z_j|\ge|\sum_{j\in J-\{k\}} z_j|\ge\sum_{j\in J-\{k\}} \Re(e^{it}z_j),$$
which shows $\Re(e^{it}z_k)\ge0$. For $k\notin J$, a similar argument shows $\Re(e^{it}z_k)\le0$. Then
$$ |\sum_{j\in J} z_j|=\sum_{j\in J} \Re(e^{it}z_j)=\sum_{j=1}^n \Re^+(e^{it}z_j). $$
The other direction is easier: for every $t\in[0,2\pi]$, let $J_t=\{j\in\{1,\dots,n\}:\Re(e^{it}z_j)\ge0\}$. Then
$$ \sum_{j=1}^n \Re^+(e^{it}z_j)=\sum_{j\in J_t} \Re(e^{it}z_j)=e^{it}\sum_{j\in J_t} \Re(z_j)\le |\sum_{j\in J_t} \Re(z_j)|. $$
