55 Find all natural numbers $n(\geqslant 2)$, such that there exist real numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying
$$
\left\{\left|a_{i}-a_{j}\right| \mid 1 \leqslant i<j \leqslant n\right\}=\left\{1,2, \cdots, \frac{n(n-1)}{2}\right\} .
$$
$55 n=2,3,4$ when, Figure 44 indicates that there exist the mentioned real numbers $a_{1}, a_{2}, \cdots, a_{n}$.
When $n \geqslant 5$, if there exist the mentioned real numbers $a_{1}, a_{2}, \cdots, a_{n}$. Then the distances between them are all integers, and each can only appear once, with the maximum distance $\frac{n(n-1)}{2} \geqslant 10$. Suppose $a_{n}-a_{1}=$ $\frac{n(n-1)}{2}$. The other numbers are all between $a_{1}$ and $a_{n}$. Since there is a difference of $\frac{n(n-1)}{2}-1$, then $a_{1}+1$ or $a_{n}-$ $1 \in\left\{a_{2}, a_{3}, \cdots, a_{n-1}\right\}=A$. Suppose $a_{2}=a_{1}+1$. Since there is a difference of $\frac{n(n-1)}{2}-2$, and $a_{2}+1 \notin A, a_{n}-1 \notin A$, then $a_{n}-2 \in A$. Set $a_{n-1}=a_{n}-2$. Consider the difference $\frac{n(n-1)}{2}$ -4. Since $a_{2}+1, a_{n-1}-2, a_{n-1}-1 \notin A$, then $a_{2}+3 \in A$. Set $a_{3}=a_{2}+3$. At this point, $a_{n-1}-a_{3}=\frac{n(n-1)}{2}-6$. Since $a_{n-1}-1, a_{n-1}-2, a_{n-1}-3 \notin A$, there does not exist a distance of $\frac{n(n-1)}{2}-5$. Therefore, $n=2, 3, 4$.