China National Team Selection Test

The maximum is $80 + 40\sqrt{3}$. Take 80 of 1's, 40 of each of $\exp(\frac{\pi}{6}i)$, $\exp(-\frac{\pi}{6}i)$, $i$, $-i$. It is straightforward to check that they satisfy (1), (2), and their sum equals $80 + 40\sqrt{3}$. We need to show this is the maximum value.

Let $z_1, z_2, \dots, z_{240}$ satisfy (1), (2). By rotation if necessary, we may assume that $S = z_1 + z_2 + \dots + z_{240}$ is a nonnegative real number; starting from $-1$ (included) along $C$, in the counterclockwise order, let the numbers be $z_1, z_2, \dots, z_{240}$.

Condition (1) indicates that for $1 \le j \le 40$, to go from $z_j$ to $z_{j+200}$ along $C$, at least an arc of length $\pi$ is needed, that is to say, there exists $2\pi \ge \alpha \ge \pi$ such that $z_{j+200} = z_j \cdot \exp(\alpha i)$. Let $z_j = (-1) \cdot \exp(\beta i)$, $\beta \in [0, 2\pi)$. From $\arg z_{j+200}$ we see $\beta + \alpha < 2\pi$, and thus
$$\operatorname{Re}(z_j + z_{j+200}) = -\cos \beta - \cos(\beta + \alpha) = -2 \cos \frac{\alpha}{2} \cos \left(\beta + \frac{\alpha}{2}\right) \le 0 \\ \left( \text{because } \frac{\alpha}{2} \in \left[\frac{\pi}{2}, \pi\right], \beta + \frac{\alpha}{2} \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \right). \text{ Summing up } 1 \le j \le 40, \text{ we} \\ \text{obtain}$$
$$\operatorname{Re}(z_1 + z_2 + \cdots + z_{40} + z_{201} + z_{202} + \cdots + z_{240}) \le 0. \quad \textcircled{1}$$

Similarly, condition (2) indicates that for $41 \le j \le 80$, to go from $z_j$ to $z_{j+120}$ along $C$, at least an arc of length $\frac{\pi}{3}$ is needed. So there exists $2\pi \ge \alpha \ge \frac{\pi}{3}$, such that $z_{j+120} = z_j \cdot \exp(\alpha i)$. Let $z_j = (-1) \cdot \exp(\beta i)$, $\beta \in [0, 2\pi)$. From $\arg z_{j+120}$ we see $\beta + \alpha < 2\pi$, and thus
$$\operatorname{Re}(z_j + z_{j+120}) = -\cos \beta - \cos(\beta + \alpha) = -2 \cos \frac{\alpha}{2} \cos \left(\beta + \frac{\alpha}{2}\right).$$
If $\alpha \ge \pi$, the above quantity is non-positive; if $\alpha \in [\frac{\pi}{3}, \pi)$, then
$$\left| 2 \cos \frac{\alpha}{2} \cos \left( \beta + \frac{\alpha}{2} \right) \right| \le 2 \times \cos \frac{\pi}{6} \times 1 = \sqrt{3},$$
so the real part is $\le \sqrt{3}$. Summing up for $41 \le j \le 80$, we obtain
$$\operatorname{Re}(z_{41} + z_{42} + \cdots + z_{80} + z_{161} + z_{162} + \cdots + z_{200}) \le 40\sqrt{3}. \quad \textcircled{2}$$

The inequalities (1) and (2) lead to an estimate of the sum
$$\begin{aligned} & |z_1 + z_2 + \cdots + z_{240}| \\ &= \operatorname{Re} \left( \sum_{j=1}^{40} (z_j + z_{200+j}) \right) + \operatorname{Re} \left( \sum_{j=41}^{80} (z_j + z_{120+j}) \right) + \operatorname{Re} \left( \sum_{j=81}^{120} z_j \right) \\ &\le 0 + 40\sqrt{3} + 80 = 80 + 40\sqrt{3}. \end{aligned}$$

Here, the maximum is attained only when the numbers are: 80 of 1's, 40 of each of $\exp(\frac{\pi}{6}i)$, $\exp(-\frac{\pi}{6}i)$, $i$, $-i$ (or by any rotation). $\square$