C42 (36-3, Japan) 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 the following two conditions:
(1) No three points of $A_{1}, A_{2}, \cdots, A_{n}$ are collinear;
(2) For each triple $i, j, k(1 \leqslant i<j<k \leqslant n), \triangle A_{i} A_{j} A_{k}$ has area equal to $r_{i}+r_{j}+r_{k}$.
The solution is that only $n=4$ satisfies the problem.
First, we prove the following two lemmas.
Lemma 1 For $\triangle A_{1} A_{2} A_{3}$, it is impossible to have two points $A_{4}, A_{5}$ such that $A_{1} A_{2} A_{3} A_{4}$ and $A_{1} A_{2} A_{3} A_{5}$ are both convex quadrilaterals and are on the same side of the lines $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{1}$.
If not, suppose there exist such points $A_{4}, A_{5}$. Then
$$
S_{\triangle A_{1} A_{2} A_{3}}+S_{\triangle A_{1} A_{3} A_{4}}=S_{A_{1} A_{2} A_{3} A_{4}}=S_{\triangle A_{1} A_{2} A_{4}}+S_{\triangle A_{2} A_{3} A_{4}},
$$

Thus,
$$
r_{1}+r_{2}+r_{3}+r_{1}+r_{3}+r_{4}=r_{1}+r_{2}+r_{4}+r_{2}+r_{3}+r_{4} \text {, }
$$

Hence,
$$
\begin{array}{l}
r_{1}+r_{3}=r_{2}+r_{4} . \\
r_{1}+r_{3}=r_{2}+r_{5} .
\end{array}
$$

Similarly,
Thus,
$$
r_{4}=r_{5} \text {. }
$$

Therefore, $S_{\triangle A_{1} A_{2} A_{4}}=S_{\triangle A_{1} A_{2} A_{5}}$, so $A_{4} A_{5} \parallel A_{1} A_{2}$, and similarly, $A_{4} A_{5} \parallel A_{2} A_{3}$, which contradicts the fact that $A_{1}, A_{2}, A_{3}$ are not collinear.

Lemma 2 If there exists a $\triangle A_{1} A_{2} A_{3}$, then it is impossible to have two known points inside it.
If $\triangle A_{1} A_{2} A_{3}$ contains two points $A_{4}, A_{5}$, then by
$$
S_{\triangle A_{1} A_{2} A_{3}}=S_{\triangle A_{1} A_{2} A_{4}}+S_{\triangle A_{1} A_{3} A_{4}}+S_{\triangle A_{2} A_{3} A_{4}},
$$

We get
$$
r_{1}+r_{2}+r_{3}=r_{1}+r_{2}+r_{4}+r_{1}+r_{3}+r_{4}+r_{2}+r_{3}+r_{1} \text {, }
$$

Thus,
$$
\begin{array}{l}
r_{4}=-\frac{r_{1}+r_{2}+r_{3}}{3} . \\
r_{5}=-\frac{r_{1}+r_{2}+r_{3}}{3} .
\end{array}
$$

Similarly,
Thus, $r_{4}=r_{5}$, and as shown above, it can be deduced that $A_{1}, A_{2}, A_{3}$ are collinear, which is a contradiction.
Consider the convex hull of $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If the number of sides of the convex hull is greater than 4, assume $A_{1} A_{2} A_{3} A_{4} A_{5}$ is a convex pentagon. For $\triangle A_{1} A_{2} A_{3}$, both $A_{1} A_{2} A_{3} A_{4}$ and $A_{1} A_{2} A_{3} A_{5}$ are convex quadrilaterals. By Lemma 1, this is impossible; if the convex hull is a quadrilateral $A_{1} A_{2} A_{3} A_{4}$ and point $A_{5}$ is inside the convex quadrilateral $A_{1} A_{2} A_{3} A_{4}$, assume point $A_{5}$ is inside $\triangle A_{1} A_{3} A_{4}$. Then for $\triangle A_{1} A_{2} A_{3}$, by Lemma 1, this is also impossible; if the convex hull is a triangle, by Lemma 2, there can be at most one point inside it. In summary, $n \leqslant 4$.

Next, we provide an example to show that $n=4$ is possible. Let $A_{1} A_{2} A_{3} A_{4}$ be a unit square, and $r_{1}=r_{2}=r_{3}=r_{4}=\frac{1}{6}$. Then,
$$
S_{\triangle A_{i} A_{j} A_{k}}=\frac{1}{2}=r_{i}+r_{j}+r_{k}(1 \leqslant i<j<k \leqslant 4),
$$

satisfying conditions (1) and (2).
In conclusion, only $n=4$ satisfies the problem.