This question from nearly a week ago seems never to have been resolved favorably, so I will attempt to give a brief exposition of its solution.
First, we must be clear on what constitutes a bus ticket. From the wording of the problem, it seems to me that any string of six digits, from 000000 to 999999 is a valid ticket, since this would enable the bus system to have an ordering of tickets. In other words, tickets beginning with one or more zeroes are perfectly acceptable.
If we agree to this definition of a ticket, the problem becomes straightforward. Let an arbitrary lucky bus ticket be denoted by its six-digits as abcdef. Write the value of the ticket as follows:
1000abc + def
where we think of abc and def as normal three-digit integers. Now, if abcdef is lucky, then so is defabc. Note that if we add together the values of the two lucky tickets abcdef and defabc, we get
1001(abc + def),
and since 1001 is divisible by 13, then any pair of lucky tickets of this type sum to a multiple of 13. The last thing to note is that any ticket of the form abcabc will only occur once in the sum of lucky tickets, not as a pair of tickets. However, tickets of this form are already divisible by 13, since they have the value
1000abc + abc = 1001abc.
So, the sum of all the lucky bus tickets must be divisible by 13.
Since 1001 = 7·11·13, the sum of all the bus tickets is also divisible by 7 and 11, we could say that since 7 and 11 are (maybe) lucky numbers, these outweigh the unlucky influence of 13!
I hope this explanation is clear.