Example 6 Determine all 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:
(1) No three points among $A_{1}, A_{2}, \cdots, A_{n}$ are collinear;
(2) For each triple $i, j, k(1 \leqslant j<k \leqslant n), \triangle A_{i} A_{j} A_{k}$ has an area equal to $r_{i}+r_{j}+r_{k}$.
Analysis Consider various possible cases of the convex hull, discuss and explore, guess that $n \geqslant 5$ is impossible. To illustrate this, it is sufficient to show that $n=5$ is not possible.
Solution Let the area of $\triangle A_{i} A_{j} A_{k}$ be denoted as $S_{i j k}$.
Lemma 1 If the vertices of a certain convex quadrilateral are 4 points $A_{1}, A_{2}, A_{3}$, and $A_{4}$ that satisfy the conditions, then there exist corresponding real numbers $r_{1}, r_{2}, r_{3}, r_{4}$ that satisfy: $r_{1}+r_{3}=r_{2}+r_{4}$.

Proof By dissecting the convex quadrilateral $A_{1} A_{2} A_{3} A_{4}$ into two triangles in two different ways, calculate the areas of the 4 triangles, and list 4 equations according to the problem conditions to solve for 4 real numbers $r_{1}, r_{2}, r_{3}, r_{4}$, which satisfy $r_{1}+r_{3}=r_{2}+r_{4}$.

Lemma 2 If 5 points $A_{1}, A_{2}, A_{3}, A_{1}, A_{5}$ satisfy the problem conditions, then at least two of the real numbers $r_{1}, r_{2}, r_{3}, r_{\mathrm{i}}, r_{\mathrm{B}}$ that satisfy the conditions are equal.
Proof Classify and discuss based on the convex hull of these 5 points.
(1) If the convex hull of these 5 points is a pentagon, denoted as $A_{1} A_{2} A_{3} A_{4} A_{5}$, then $A_{1} A_{2} A_{3} A_{5}$ and $A_{1} A_{2} A_{4} A_{5}$ are both convex quadrilaterals, as shown in Figure 23-9. By Lemma 1, there exist corresponding real numbers $r_{1}, r_{2}, r_{3}, r_{4}, r_{\mathrm{3}}$ that satisfy:
$$
r_{1}+r_{3}=r_{2}+r_{5}=r_{1}+r_{4} \Rightarrow r_{3}=r_{4} .
$$
(2) If the convex hull of these 5 points is a quadrilateral $A_{1} A_{2} A_{3} A_{4}$, assume $A_{5}$ is inside $\triangle A_{1} A_{2} A_{3}$, as shown in Figure 23-10. Then $A_{1} A_{2} A_{3} A_{4}$ and $A_{1} A_{5} A_{3} A_{1}$ are both convex quadrilaterals. Similarly to (1), we have $r_{2}=r_{3}$.
(3) If the convex hull of these 5 points is $\triangle A_{1} A_{2} A_{3}$, and $A_{1}, A_{5}$ are inside the triangle, as shown in Figure 23-11, then $S_{124}+S_{234}+S_{314}=S_{125}+S_{235}+S_{315} \Rightarrow r_{4}=r_{5}$.
Thus, Lemma 2 is proven.
Below, we will show that $n \geqslant 5$ is impossible.
If not, assume 5 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ satisfy the problem conditions.
By Lemma 2, among the corresponding real numbers $r_{1}, r_{2}, r_{3}, r_{4}, r_{3}$, assume $r_{1}=r_{2}$.
It is easy to see that $A_{1}, A_{2}$ must be on the same side of $A_{3} A_{4}, A_{4} A_{5}, A_{5} A_{3}$, assume $A_{1}, A_{2}$ are on the same side of $A_{3} A_{4}$. Since $S_{134}=S_{234}$, we have $A_{1} A_{2} \parallel A_{3} A_{4} \Rightarrow S_{123}=S_{124} \Rightarrow r_{3}=r_{1}$.

By $S_{145}=S_{245}$, and $A_{1} A_{2}$ is not parallel to $A_{4} A_{5}$ (otherwise $A_{3}, A_{4}, A_{5}$ are collinear), we know that $A_{4} A_{5}$ bisects segment $A_{1} A_{2}$;
By $S_{135}=S_{225}$, and $A_{1} A_{2}$ is not parallel to $A_{3} A_{5}$, we know that $A_{3} A_{5}$ bisects segment $A_{1} A_{2}$.
Thus, $A_{5}$ is the midpoint of segment $A_{1} A_{2}$, which means $A_{1}, A_{5}, A_{2}$ are collinear, a contradiction!
Therefore, $n \geqslant 5$ does not hold. Below, we provide an example to show that $n=4$ is possible.
Take $A_{1}, A_{2}, A_{3}, A_{1}$ as the four vertices of a unit square, and let $r_{1}=r_{2}=r_{3}=r_{4}=\frac{1}{6}$.
In summary, $n=4$ is the only possible case.
/Comment The two lemmas are just two properties discovered during the exploration process. Guessing that $n \leqslant 4$ is the breakthrough for solving the problem! The essence of this problem is to describe the area and positional relationships in plane geometry using real numbers.