$10 \cdot 86$ Find all integers $n>3$, such that there exist $n$ points $A_{1}, A_{2}, \cdots$, $A_{n}$ and real numbers $r_{1}, r_{2}, \cdots, r_{n}$ in the plane, satisfying the following two conditions:
(1) No three points among $A_{1}, A_{2}, \cdots, A_{n}$ are collinear;
(2) For each triplet $\left\{A_{i}, A_{j}, A_{k}\right\}, 1 \leqslant i<j<k \leqslant n, \triangle A_{i} A_{j} A_{k}$ has an area $S_{i j k}=r_{i}+r_{j}+r_{k}$.
[Solution] For convenience, we first provide two lemmas.

Lemma 1 If the 4 vertices of a convex quadrilateral are the given points $A_{i}, A_{j}, A_{k}$, and $A_{h}$ in sequence, then the corresponding 4 real numbers satisfy the relation
$$
r_{i}+r_{k}=r_{j}+r_{h}.
$$

Proof of Lemma 1: Connect the diagonals $A_{i} A_{k}$ and $A_{j} A_{h}$, then we have
$$
S_{i j k}+S_{i k h}=S_{i j k}+S_{j k h}.
$$

Therefore,
$$
2 r_{i}+r_{j}+2 r_{k}+r_{h}=r_{i}+2 r_{j}+r_{k}+2 r_{h}.
$$

From this, we get $r_{i}+r_{k}=r_{j}+r_{h}$.

Lemma 2 If there are 5 given points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ satisfying the conditions of the problem, then among the corresponding 5 real numbers $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$, at least two numbers are equal.

Proof of Lemma 2: Consider the convex hull of the 5 points.
(i) Suppose the convex hull is a pentagon. Without loss of generality, assume its vertices are $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ in sequence. 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
$$
r_{2}+r_{4}=r_{1}+r_{3}=r_{2}+r_{5}.
$$

Thus, we get $r_{4}=r_{5}$.
(ii) Suppose the convex hull is a quadrilateral. Without loss of generality, assume point $A_{5}$ is inside the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ and is in the interior of $\triangle A_{2} A_{3} A_{4}$. Then the quadrilateral $A_{1} A_{2} A_{5} A_{4}$ is also a convex quadrilateral. By Lemma 1, we have
$$
r_{1}+r_{3}=r_{2}+r_{4}=r_{1}+r_{5}.
$$

Thus, we get $r_{3}=r_{5}$.
(iii) Suppose the convex hull is $\triangle A_{1} A_{2} A_{3}$, i.e., $A_{4}, A_{5}$ are inside $\triangle A_{1} A_{2} A_{3}$. Then we have
$$
S_{124}+S_{234}+S_{341}=S_{125}+S_{235}+S_{351}.
$$

Therefore,
$$
2 r_{1}+2 r_{2}+2 r_{3}+3 r_{4}=2 r_{1}+2 r_{2}+2 r_{3}+3 r_{5}.
$$

Thus, we get $r_{4}=r_{5}$. This completes the proof of Lemma 2.

By Lemma 1 and Lemma 2, we can prove that all integers $n \geqslant 5$ do not satisfy the requirements of the problem. If not, suppose the set of five points $\left\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\right\}$ satisfies the conditions of the problem. Then by Lemma 2, the corresponding 5 real numbers $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$ must have two numbers equal, without loss of generality, assume $r_{4}=r_{5}$.
(a) Suppose points $A_{1}, A_{2}, A_{3}$ are on the same side of the line $A_{4} A_{5}$. Then by the given conditions, we have
$$
S_{124}=S_{125}, S_{234}=S_{235}.
$$

From this, we get $A_{1} A_{2} \parallel A_{4} A_{5}, A_{2} A_{3} \parallel A_{4} A_{5}$, thus $A_{1}, A_{2}, A_{3}$ are collinear, which contradicts condition (1).
(b) Suppose $A_{1}, A_{2}$ are on the same side of the line $A_{4} A_{5}$ and $A_{3}$ is on the other side. Then we have
$$
S_{124}=S_{125}.
$$

Therefore, $A_{1} A_{2} \parallel A_{4} A_{5}$. From this, we also get
$$
S_{145}=S_{245}.
$$

Thus, $r_{1}=r_{2}$. Since $A_{3}, A_{4}, A_{5}$ are on the same side of the line $A_{1} A_{2}$, similarly, we can derive a contradiction. This proves that all $n \geqslant 5$ cannot satisfy the requirements of the problem.

When $n=4$, take the 4 vertices of a unit square as $A_{1}, A_{2}, A_{3}, A_{4}$, and take $r_{1}=r_{2}=r_{3}=r_{4}=\frac{1}{6}$. It is easy to verify that these 4 points and 4 real numbers satisfy the requirements of the problem.
In summary, the only integer $n$ that satisfies the requirements of the problem is $n=4$.