One, (40 points) For a simple 300-gon in the plane (non-self-intersecting), find the maximum number of right angles among all its interior angles.
Let $\theta_{i}(i=1,2, \cdots, 300)$ be the interior angles, and $k$ of them are right angles. Without loss of generality, let $\theta_{1}=\theta_{2}=\cdots=\theta_{k}=\frac{\pi}{2}$.

Since $\theta_{i}k)$, and the sum of the interior angles of a 300-sided polygon is $298 \pi$, we have:
$$
\begin{array}{l}
298 \pi=\sum_{i=1}^{300} \theta_{i}=\frac{\pi}{2} \cdot k+\sum_{i=k+1}^{300} \theta_{i} \\
<\frac{\pi}{2} \cdot k+2 \pi(300-k) \\
\Rightarrow 600-4<k+1200-4 k \\
\Rightarrow 3 k<604 \Rightarrow k \leqslant 201 .
\end{array}
$$

Consider a "palm" with 100 "fingers" (as shown in Figure 4), where each finger has two right angles, plus the angle $\angle A_{1} O B_{100}$ of the palm is a right angle, making a total of 201 right angles. Therefore, the maximum number of right angles is 201.