Example 5. A parade team whose number of people is a multiple of 5 and no less than 1000, if lined up in rows of 4, is short of 3 people; if lined up in rows of 3, is short of 2 people; if lined up in rows of 2, is short of 1 person. Find the minimum number of people in this parade team.
Analysis: Given that the total number of people in the parade is a multiple of 5, let the total number of people be $5n$. "When arranged in rows of 4, there are 3 people short," which can be understood as "1 person more" from the opposite perspective. Similarly, when arranged in rows of 3 or 2, it can also be understood as "1 person more." Clearly, the problem can be transformed into a congruence problem, where $5n$ leaves a remainder of 1 when divided by 4, 3, and 2, i.e., $5n-1$ is a multiple of 12. Given that the total number of people is no less than 1000, the solution to the problem can be found.

Let the total number of people in the parade be $5n (n \in N)$, i.e., $5n-1$ is a common multiple of 4, 3, and 2. Thus, we can let $5n-1=12m (m \in N)$, which gives $n=\frac{12m+1}{5}$.

To minimize the number of people in the parade, the $m$ in equation (A) should be the smallest positive integer and a multiple of 5, which is 2. Therefore, we can let $m=5p+2 (p \in N)$. This gives:
$$
n=\frac{1}{5}[12(5p+2)+1]=12p+5,
$$

i.e., $5n \geqslant 60p+25$.
$$
\therefore 60p+25 \geqslant 1000, \quad p \geqslant 16 \frac{1}{4}.
$$

Taking $p=17$, substituting into equation (B) gives $5n=60 \times 17+25=1045$.
Therefore, the minimum total number of people in the parade is 1045.