Solution:

Let $X = \{P_1, P_2, \ldots, P_n\}$. For each vector $AB$ in $S$, let $\vec{v}_{AB}$ denote the vector from $A$ to $B$.

The sum of all vectors in $S$ is:
$$\sum_{AB \in S} \vec{v}_{AB} = \sum_{AB \in S} (\vec{OB} - \vec{OA})$$
where $O$ is the origin and $\vec{OA}$ is the position vector of $A$.

This can be rewritten as:
$$\sum_{AB \in S} \vec{OB} - \sum_{AB \in S} \vec{OA}$$

For each point $A$, let $d^+_A$ be the number of vectors in $S$ starting at $A$ (i.e., the number of $AB$ in $S$ for fixed $A$), and $d^-_A$ be the number of vectors in $S$ ending at $A$ (i.e., the number of $CA$ in $S$ for fixed $A$).

Then:
$$\sum_{AB \in S} \vec{OA} = \sum_{A \in X} d^+_A \vec{OA}$$
$$\sum_{AB \in S} \vec{OB} = \sum_{B \in X} d^-_B \vec{OB}$$

So the total sum is:
$$\sum_{B \in X} d^-_B \vec{OB} - \sum_{A \in X} d^+_A \vec{OA} = \sum_{A \in X} (d^-_A - d^+_A) \vec{OA}$$

But by the given, for every $A$, $d^+_A = d^-_A$, so $d^-_A - d^+_A = 0$ for all $A$.

Therefore, the sum is zero.