$13 \cdot 81$ Suppose in a certain forest there are $n \geqslant 3$ starling nests, and the distances between them are all different. In each nest, there is one starling. At some moment, some of the starlings fly from their nests to other nests. If among the flying starlings, the distance between one pair of starlings is less than the distance between another pair of starlings (a starling can form a "pair" with any other starling), then after landing, the distance between the first pair of starlings is always greater than the distance between the second pair of starlings. For what value of $n$ is this possible?
[Solution] When $n \geqslant 4$, suppose that at a certain time, 4 starlings fly up, and their 4 nests are $A_{1}, A_{2}, A_{3}, A_{4}$, while the 4 nests they land on are $A^{\prime}, A_{1}^{\prime}{ }_{2}, A^{\prime}{ }_{3}, A^{\prime}{ }_{4}$, and $A_{1}^{\prime}, A_{2}^{\prime}, A_{3}^{\prime}, A_{4}^{\prime}$ is a permutation of $A_{1}, A_{2}, A_{3}, A_{4}$. Since the distances between any two pairs of birds must be reciprocal, we have
$$
\begin{array}{l}
\left\{A_{1}^{\prime}, A_{2}^{\prime}\right\}=\left\{A_{3}, A_{4}\right\},\left\{A_{1}^{\prime}, A_{3}^{\prime}\right\}=\left\{A_{2}, A_{4}\right\}, \\
\left\{A_{1}^{\prime}, A_{4}^{\prime}\right\}=\left\{A_{2}, A_{3}\right\} .
\end{array}
$$

Thus, we have
$$
A_{1}^{\prime} \in\left\{A_{3}, A_{4}\right\} \cap\left\{A_{2}, A_{4}\right\} \cap\left\{A_{2}, A_{3}\right\}=\varnothing,
$$

which is impossible. This contradiction shows that $n \geqslant 4$ does not satisfy the requirements of the problem. However, $n=3$ clearly satisfies the requirements of the problem.