Example 6: Using the digits $0,1,2,3,4,5,6$, several seven-digit numbers without repeated digits can be formed, among which there are multiples of 55. Try to find the largest and smallest numbers among the seven-digit multiples of 55.
Let the seven-digit number be $N=\overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, and $a_{6} \neq 0, a_{i}(i=0,1, \cdots, 5)$ can take any digit from 0 to 6.
$$
\text { Let } A=a_{6}+a_{4}+a_{2}+a_{0}, B=a_{5}+a_{3}+a_{1} \text {, }
$$

Then $A+B=21$.
A seven-digit number that can be divided by 55 must be divisible by both 11 and 5, so $|A-B|=11 k$ (where $k$ is zero or a natural number).
From (3), we get $|A-B|<21, A-B \neq 0$, thus $k=1$, i.e., $|A-B|=11$.
From (3) and (4), we know that one of $A, B$ is 16, and the other is $\frac{21-11}{2}=5$. Since the sum of the smallest four numbers among 0,1, $2,3,4,5,6$ is 6, we have $A=16, B=5$. Therefore, the three digits that make up $B$ can only be $0,1,4$ or $0,2,3$ in two cases, i.e.,
$$
B=0+1+4, A=2+3+5+6 \text {, or } B=0+2+3, A=1+4+5+6 \text {. }
$$

Since 0 is not in the last position of the seven-digit number, the last digit must be 5, i.e., $a_{0}=5$, so the smallest seven-digit number is 1042635, and the largest seven-digit number is 6431205.