Find the smallest positive real number $r$ with the following property:  For every choice of $2023$ unit vectors $v_1,v_2, \dots ,v_{2023} \in \mathbb{R}^2$, a point $p$ can be found in the plane such that for each subset $S$ of $\{1,2, \dots , 2023\}$, the sum
$$\sum_{i \in S} v_i$$
lies inside the disc $\{x \in \mathbb{R}^2 : ||x-p|| \leq r\}$.
1. **Restate the problem in general terms:**
   We need to find the smallest positive real number \( r \) such that for any set of \( N \) unit vectors \( v_1, v_2, \ldots, v_N \in \mathbb{R}^2 \), there exists a point \( p \) in the plane such that for every subset \( S \) of \(\{1, 2, \ldots, N\}\), the sum \( \sum_{i \in S} v_i \) lies within the disc \( \{x \in \mathbb{R}^2 : \|x - p\| \leq r\} \).

2. **Propose a candidate for \( p \):**
   Let \( p = \frac{1}{2} \sum_{i=1}^N v_i \). This choice of \( p \) is motivated by the idea that it is the centroid of the vectors scaled by \( \frac{1}{2} \).

3. **Verify that this \( p \) works:**
   For any subset \( S \subseteq \{1, 2, \ldots, N\} \), consider the sum \( \sum_{i \in S} v_i \). We need to show that this sum lies within the disc centered at \( p \) with radius \( r \).

   Using the triangle inequality, we have:
   \[
   \left\| \sum_{i \in S} v_i - p \right\| = \left\| \sum_{i \in S} v_i - \frac{1}{2} \sum_{i=1}^N v_i \right\|
   \]
   This can be rewritten as:
   \[
   \left\| \sum_{i \in S} v_i - \frac{1}{2} \sum_{i=1}^N v_i \right\| = \left\| \sum_{i \in S} v_i - \frac{1}{2} \left( \sum_{i \in S} v_i + \sum_{i \notin S} v_i \right) \right\|
   \]
   Simplifying further:
   \[
   \left\| \sum_{i \in S} v_i - \frac{1}{2} \sum_{i \in S} v_i - \frac{1}{2} \sum_{i \notin S} v_i \right\| = \left\| \frac{1}{2} \sum_{i \in S} v_i - \frac{1}{2} \sum_{i \notin S} v_i \right\|
   \]
   Using the triangle inequality again:
   \[
   \left\| \frac{1}{2} \sum_{i \in S} v_i - \frac{1}{2} \sum_{i \notin S} v_i \right\| \leq \frac{1}{2} \left\| \sum_{i \in S} v_i \right\| + \frac{1}{2} \left\| \sum_{i \notin S} v_i \right\|
   \]
   Since each \( v_i \) is a unit vector, the maximum possible value of \( \left\| \sum_{i \in S} v_i \right\| \) is \( |S| \) and similarly for \( \left\| \sum_{i \notin S} v_i \right\| \). Therefore:
   \[
   \left\| \frac{1}{2} \sum_{i \in S} v_i - \frac{1}{2} \sum_{i \notin S} v_i \right\| \leq \frac{1}{2} N
   \]

4. **Show that \( r = \frac{N}{2} \) is necessary:**
   Consider the case where all \( v_i \) are the same unit vector, say \( v_i = e_1 \) for all \( i \). Then:
   \[
   \sum_{i \in S} v_i = |S| e_1
   \]
   If \( S = \emptyset \), then \( \sum_{i \in S} v_i = 0 \). If \( S = \{1, 2, \ldots, N\} \), then \( \sum_{i \in S} v_i = N e_1 \). The distance between these two points is \( N \), and since \( p = \frac{N}{2} e_1 \), we have:
   \[
   N = \left\| N e_1 - 0 \right\| \leq \left\| N e_1 - p \right\| + \left\| p - 0 \right\| = r + r = 2r
   \]
   Thus, \( r \geq \frac{N}{2} \).

5. **Conclusion:**
   Combining both parts, we conclude that the smallest \( r \) that satisfies the condition is \( \frac{N}{2} \).

The final answer is \( \boxed{\frac{N}{2}} \)