$10 \cdot 17$ A bus ticket number is a six-digit number. If the sum of the first three digits equals the sum of the last three digits, the ticket is called a "lucky ticket." How many consecutive ticket numbers must be bought from the ticket office to ensure that at least one of them is a "lucky ticket"?
(St. Petersburg Mathematical Olympiad, 1987)
[Solution] Buying 1001 consecutive tickets ensures that at least one of them is a "lucky ticket".

In fact, if the number of consecutive tickets purchased does not exceed 1000, then when the smallest number of the tickets purchased is 000001, there will be no "lucky ticket" among them.

If 1001 consecutive tickets are bought, and let the smallest number be $\overline{x y z a b c}$, where $x, y, z, a, b, c \in\{0,1, \cdots, 9\}$. Then, when $z \neq 9$, the largest number is $\overline{x y}(z+1) a b c$. In this case, when $\overline{a b c} \leqslant \overline{x y z}$, $\overline{x y z x y z}$ is among the tickets purchased, and when $\overline{a b c}>\overline{x y z}$, $\overline{x y(z+1) x y(z+1)}$ is among the tickets purchased.

If $z=9, y \neq 9$, then when the smallest number is $x y z a b c$, the largest number is $\overline{x(y+1) 0 a b c}$, and the same discussion can be made.
If $\overline{x y z}=999$, then $\overline{999999}$ must be among the tickets purchased.
From the above, for 1001 consecutive tickets, there must be one "lucky ticket".