Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how many computers should be transferred to the neighboring school: No.1 to No.2, No.2 to No.3, No.3 to No.4, No.4 to No.5, and No.5 to No.1? If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, what arrangement should be made?
Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x_{2}=11+x_{2}-x_{3}=3+x_{3}-x_{4} \\
=14+x_{4}-x_{5}=15+x_{5}-x_{1}=10 .
\end{array}
$$

Thus, we have $x_{2}=x_{1}-3, x_{3}=x_{1}-2$,
$$
x_{4}=x_{1}-9, x_{5}=x_{1}-5 \text {. }
$$

According to the problem, we have
$$
\begin{aligned}
y= & \left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|+\left|x_{4}\right|+\left|x_{5}\right| \\
= & \left|x_{1}\right|+\left|x_{1}-3\right|+\left|x_{1}-2\right|+\left|x_{1}-9\right|+ \\
& \left|x_{1}-5\right| \\
= & \left|x_{1}\right|+\left|x_{1}-2\right|+\left|x_{1}-3\right|+\left|x_{1}-5\right|+ \\
& \left|x_{1}-9\right| .
\end{aligned}
$$

By the geometric meaning of absolute value, when $x_{1}=3$, $y$ has the minimum value, which is 12. At this time, $x_{2}=0$, $x_{3}=1, x_{4}=-6, x_{5}=-2$.

Therefore, the transfer plan is: School 1 transfers 3 computers to School 2, School 3 transfers 1 computer to School 4, School 5 transfers 6 computers to School 4, and School 1 transfers 2 computers to School 5. This way, the total number of computers transferred is minimized, with the minimum number being 12.

Note: Using the geometric meaning of absolute value, the following conclusion can be explained:

Let $a_{1}, a_{2}, \cdots, a_{n}$ be the real numbers represented by points on the number line arranged in sequence.
When $n$ is odd, if $x=a_{\frac{n+1}{2}}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$

has the minimum value;
When $n$ is even, if $a_{\frac{n}{2}} \leqslant x \leqslant a_{\frac{n}{2}+1}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$

has the minimum value.