$7 \cdot 76$ A highway has 10 bus stops $A_{0}, A_{1}, A_{2}, \cdots, A_{9}$. The distance between adjacent stops is $a$ kilometers. A car departs from station $A_{0}$, travels to all stations to deliver goods. The car only stops once at each station and finally returns to the starting station $A_{0}$. Due to the need for cargo transportation, the car does not necessarily stop at stations $A_{1}, A_{2}, A_{3}, \cdots$ in sequence (for example, it can go from $A_{3} \rightarrow A_{5} \rightarrow A_{7} \rightarrow \cdots$). What is the maximum possible distance the car can travel in kilometers? What is the minimum distance? Explain your reasoning.
[Solution] From the given information, the distances from stations $A_{0}, A_{1}, A_{2}, \cdots, A_{9}$ to $A_{0}$ are $0, a$, $2 a, \cdots, 9 a$ respectively.

Let the distance from $A_{0}$ to the station where the car stops for the $n$-th time be $x_{n}\left(x_{n} \geqslant 0\right)$. Then the total distance traveled by the car is:
$$
S=\left|x_{0}-x_{1}\right|+\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{9}-x_{0}\right| .
$$

If we remove the absolute value signs from the above expression, we get that $S$ is the algebraic sum of 10 positive numbers and 10 negative numbers. Clearly, this sum is no greater than
$$
\begin{aligned}
& 2(9 a+8 a+7 a+6 a+5 a)-2(4 a+3 a+2 a+a+0) \\
= & 50 a \text { (km) }
\end{aligned}
$$

Therefore, the distance traveled by the car does not exceed $50 a$ km.
When the car travels along the following route:
$$
\begin{aligned}
A_{0} \rightarrow & A_{9} \rightarrow A_{1} \rightarrow A_{8} \rightarrow A_{2} \rightarrow A_{7} \rightarrow A_{3} \rightarrow A_{6} \rightarrow A_{4} \rightarrow A_{5} \rightarrow A_{0} \\
S= & |0-9 a|+|9 a-a|+|a-8 a|+|8 a-2 a| \\
& +|2 a-7 a|+|7 a-3 a|+|3 a-6 a|+|6 a-4 a| \\
& +|4 a-5 a|+|5 a-0| \\
= & +9 a+8 a+7 a+6 a+5 a+4 a+3 a+2 a+a+5 a \\
= & 50 a \text { (km). }
\end{aligned}
$$

Thus, the maximum distance traveled is $50 a$ km.
If the car travels back and forth between $A_{0}$ and $A_{9}$, the distance traveled is at least twice the distance between $A_{0}$ and $A_{9}$, i.e.,
$$
S \geqslant 2 \times 9 a=18 a \text { (km). }
$$

When the car travels along the following route:
$$
\begin{aligned}
A_{0} \rightarrow & A_{1} \rightarrow A_{2} \rightarrow A_{3} \rightarrow A_{4} \rightarrow A_{5} \rightarrow A_{6} \rightarrow A_{7} \rightarrow A_{8} \rightarrow A_{9} \rightarrow A_{0} \\
S= & |0-a|+|a-2 a|+|2 a-3 a|+|3 a-4 a| \\
& +|4 a-5 a|+|5 a-6 a|+|6 a-7 a|+|7 a-8 a| \\
& +|8 a-9 a|+|9 a-0| \\
= & 18 a \text { (km) }
\end{aligned}
$$

Therefore, the minimum distance traveled by the car is $18 a$ km.