The answer is (if we replace $2023$ with $N$) $\frac{N}{2}$. Indeed, if we take $p=\frac{1}{2}\sum_{i\in[N]}v_i$ and $S\subseteq [N]=\{1,...,N\}$, it follows by triangle inequality that $|x-p|=|\frac 12 \sum_{i\in S} v_i-\frac 12\sum_{i\in [N]\setminus S} v_i|\leq \frac 12\sum_{i\in [N]}|v_i|=\frac N2$. However, if we take all $v_i$ to be equal to some equal versor, if $S=\{\}$ then $x=0$, while if $S=[N]$ then $|x|=N$. Again by triangle inequality, if follows that $r$ must be at least $N/2$, since $N=|x-0|\leq |x-p|+|p-0|\leq r+r$. Therefore, the minimal $r$ is exactly $r=\frac N2$