First, prove the following two lemmas:
Lemma 1 If the four vertices of a convex quadrilateral are four given points $A_{i}, A_{j}, A_{k}, A_{h}(1 \leqslant i<j<k<h \leqslant n)$, then $r_{i}+r_{k}=r_{j}+r_{h}$.
Proof of Lemma 1 Connect the diagonals $A_{i} A_{k}$ and $A_{j} A_{h}$, then
$$
S_{i j k}+S_{i k t}=S_{i j k}+S_{j k k}.
$$

Therefore, we have $2 r_{i}+r_{j}+2 r_{k}+r_{h}=r_{i}+2 r_{j}+2 r_{h}+r_{k}$.
From this, we get $r_{i}+r_{k}=r_{j}+r_{h}$.
Lemma 2 If there are 5 given points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ that satisfy the conditions of the problem, then at least 2 of $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$ are equal.

Proof of Lemma 2 (1) If the convex hull of the 5 points is a convex pentagon, without loss of generality, let it be $A_{1} A_{2} A_{3} A_{4} A_{5}$, then the quadrilaterals $A_{1} A_{2} A_{3} A_{4}$ and $A_{1} A_{2} A_{3} A_{5}$ are both convex quadrilaterals. By Lemma 1, we have
$$
r_{1}+r_{3}=r_{2}+r_{4}, r_{1}+r_{3}=r_{2}+r_{5}
$$

Thus, we get $r_{4}=r_{5}$.
(2) If the convex hull of the 5 points is a convex quadrilateral, without loss of generality, let $A_{5}$ be inside the convex quadrilateral $A_{1} A_{2} A_{3} A_{4}$ and in the interior of $\triangle A_{1} A_{3} A_{4}$, then the quadrilateral $A_{1} A_{2} A_{3} A_{5}$ is also a convex quadrilateral. Similarly to case (1), we can deduce $r_{4}=r_{5}$.
(3) If the convex hull of the 5 points is a triangle, without loss of generality, let it be $\triangle A_{1} A_{2} A_{3}$, and $A_{4}$ and $A_{5}$ are inside $\triangle A_{1} A_{2} A_{3}$, then
$$
S_{124}+S_{234}+S_{341}=S_{125}+S_{235}+S_{351}.
$$

Therefore, we have $2 r_{1}+2 r_{2}+2 r_{3}+3 r_{4}=2 r_{1}+2 r_{2}+2 r_{3}+3 r_{5}$.
Thus, we get $r_{4}=r_{5}$, and Lemma 2 is proved.
If there exist $n \geqslant 5$ points that satisfy the problem's conditions, then taking any 5 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ still satisfies the problem's conditions. By Lemma 2, we know that $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$ must have 2 that are equal, without loss of generality, let $r_{4}=r_{5}$.
(a) If $A_{1}, A_{2}, A_{3}$ are on the same side of the line $A_{4} A_{5}$, then
$$
S_{124}=S_{125}=r_{1}+r_{2}+r_{4}, S_{24}=S_{235}=r_{2}+r_{3}+r_{4}.
$$

Thus, $A_{4} A_{5} \parallel A_{1} A_{2}, A_{4} A_{5} \parallel A_{2} A_{3}$, hence $A_{1}, A_{2}, A_{3}$ are collinear, which contradicts condition (1).
(b) If $A_{1}, A_{2}$ are on the same side of the line $A_{4} A_{5}$, and $A_{3}$ is on the other side, then
$$
S_{124}=S_{125}.
$$

Therefore, $A_{4} A_{5} \parallel A_{1} A_{2}$, from which we can also get $S_{145}=S_{245}$, thus $r_{1}=r_{2}$. Since $A_{3}, A_{4}, A_{5}$ are on the same side of $A_{1} A_{2}$, similarly to (a), it leads to a contradiction. This proves that $n \geqslant 5$ does not satisfy the requirements.

When $n=4$, take the four vertices of a unit square as $A_{1}, A_{2}, A_{3}, A_{4}$, and let $r_{1}=r_{2}=r_{3}=r_{4}=\frac{1}{6}$. It is easy to verify that these 4 points and 4 real numbers satisfy the problem's requirements.
In summary, the only positive integer that satisfies the problem's requirements is $n=4$.
Example 3 Find all positive integers $n>3$, such that there exist $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in the plane and real numbers $r_{1}$, $r_{2}, \cdots, r_{n}$ satisfying the following conditions:
(1) No three points of $A_{1}, A_{2}, \cdots, A_{n}$ are collinear;
(2) For every triple $\left\{A_{i}, A_{j}, A_{k}\right\}(1 \leqslant i<j<k \leqslant n)$, the area $S_{i j k}$ of $\triangle A_{i} A_{j} A_{k}$ is $S_{i j k}=r_{i}+r_{j}+$ $r_{k}$.
(1995 36th IMO Problem)