4. Given that $\overline{73 a b c 6}$ is divisible by 56 $(b<4)$, and $a$ leaves the same remainder when divided by 40, 61, and 810, find all six-digit numbers that satisfy the requirements.
【Analysis】Given $40 \equiv 61 \equiv 810(\bmod a), a|61-40 \Rightarrow a| 21, a|810-61 \Rightarrow a| 749$. $a$ is a common divisor of 21 and 749, so $a=7$ or $a=1$.
(1) When $a=7$, $56|\overline{737 b c 6} \Rightarrow 8| \overline{737 b c 6}, 7 \mid \overline{737 b c 6}$

First satisfy 8, then verify 7. When $b=0,2$, $c=1,5,9$; when $b=1,3$, $c=3,7$; by the divisibility rule of 7, verify and enumerate to get $737016 ; 737296$
(2) When $a=1$, similarly verify that 731136 meets the requirement.

In summary, all six-digit numbers that meet the requirements are: 731136, 737016, 737296.
【Answer】 $731136,737016,737296$
$S=a_{1}+a_{2}+\cdots \cdots+a_{n}$. After $n$ times, the total number of cards all students have is $n S$, because the numbers Xiao Ming takes each time are different, so the sum of the numbers Xiao Ming takes is $S$. Considering the extreme case, if Xiao Ming is the first, because the total of each person is different, the total of all people is less than $n$ times the number $S$ of Xiao Ming, which is a contradiction.
Construct an example for the second place:
The total of all children except Xiao Ming is $n S-S=2139$, i.e., $(n-1) S=2139=3 \times 23 \times 31$.
When $n=4$, $S=713$, the 4 cards can be $1,2,3,707$.
The following distribution makes Xiao Ming the second place.
Therefore, the highest rank Xiao Ming can achieve is 2nd place.
【Answer】2