$7 \cdot 91$ On a tape, bus tickets are printed with numbers from 000000 to 999999, and then tickets whose even-positioned digits sum equals the sum of the odd-positioned digits are painted blue. Find the maximum difference between the numbers of two adjacent blue tickets.
[Solution] The maximum value of the difference in numbers between adjacent blue tickets is 990.
First, we prove that there are no blue tickets between the two blue tickets 908919 and 909909, which have a number difference of 990. If otherwise, suppose there is a blue ticket $\overline{a b c d e f}$ between them, i.e.,
$$
908919 < \overline{a b c d e f} < 909909
$$
Since $a, b, c, d, e, f \leqslant 9$, we have $f - e \leqslant 7$, which is a contradiction. This shows that there are no blue tickets between 908919 and 909909, hence the maximum value sought is no less than 990.

On the other hand, consider all tickets with numbers of the form $\overline{a b c a b c}$. Clearly, such tickets are all blue tickets and they form an arithmetic sequence with a common difference of 1001. Since the numbers of blue tickets are all multiples of 11, to prove that the difference in numbers between any two adjacent blue tickets does not exceed 990, it suffices to prove the following lemma:

Lemma: For any blue ticket number $\overline{a b c a b c} < 999999$, there is at least one blue ticket number between $\overline{a b c a b c}$ and $\overline{a b c a b c} + 1001$.
Proof of the lemma: We consider 4 cases separately as follows:
(1) If $c \neq 9, b \neq 9$, then $\overline{a b c a b c} + 11$ is a blue ticket;
(2) If $c = 9, b = 9$, then $a \neq 9$, so $\overline{a b c a b c} + 11$ is a blue ticket;
(3) If $c = 9, b \neq 9$, then $\overline{a b c a b c} + 110$ is a blue ticket;
(4) If $c \neq 9, b = 9$, then $\overline{a b c a b c} + 1001 - 11$ is a blue ticket.
In summary, the maximum value of the difference in numbers between two adjacent blue tickets is 990.