Find the smallest positive real number $r$ with the following property: For every choice of 2023 unit vectors $\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{2023} \in \mathbb{R}^{2}$, a point $\mathbf{p}$ can be found in the plane such that for each subset $S$ of $\{1,2, \ldots, 2023\}$, the sum
$$
\sum_{i \in S} \mathbf{v}_{i}
$$
lies inside the disc $\left\{\mathbf{x} \in \mathbb{R}^{2}:\|\mathbf{x}-\mathbf{p}\| \leq r\right\}$.
We will show that the answer is $r=\frac{2023}{2}$.
Let $\mathbf{p}=\frac{1}{2} \sum_{i=1}^{2023} \mathbf{v}_{i}$. If $S \subseteq\{1,2, \ldots, 2023\}$ then
$$
\left\|\left(\sum_{i \in S} \mathbf{v}_{i}\right)-\mathbf{p}\right\|=\frac{1}{2}\left\|\sum_{i \in S} \mathbf{v}_{i}-\sum_{i \notin S} \mathbf{v}_{i}\right\| \leq \frac{1}{2}\left(\sum_{i \in S}\left\|\mathbf{v}_{i}\right\|+\sum_{i \notin S}\left\|-\mathbf{v}_{i}\right\|\right)=\frac{1}{2} \times 2023
$$
by the triangle inequality. Therefore the sum $\sum_{i \in S} \mathbf{v}_{i}$ lies within the disc centred at $\mathbf{p}$ with radius $\frac{2023}{2}$.
On the other hand, if $\mathbf{v}_{1}=\mathbf{v}_{2}=\cdots=\mathbf{v}_{2023}$ then the sums corresponding to $S=\emptyset$ and $S=\{1,2, \ldots, 2023\}$ are distance 2023 apart, so cannot both lie in any disc of radius less than $\frac{2023}{2}$. Therefore $\frac{2023}{2}$ is the optimal value of $r$.