Preface
The following problem, I suppose, most of you have read in recreational math quiz books. It is stated as following:
On a stick, there are $4n+1$ ants, with $2n$ ants at one pole of the stick and $2n+1$ at the other. The ants line up, and starts to move toward the other side of the stick. Now if there is no obstacle, then the ants just move forward. If two ants meets, they change their direction. If an ant reaches an end, it moves back.
Prove that, there is a time, when all the ants are at the opposite end of where they start.
Now, it comes to me that, if we differentiate the ants, so ant $A1$ is different from $A2$, and therefore how they behave upon meeting each other is also different.
Problem
Let there be $4n+1$ people, standing at two ends of a road. One end has $2n$ people and the other has $2n+1$ holding a card from $1$ to $4n+1$. They position randomly.
If $a > b$, then $b$ turns around and moves before $a$ in $a$'s direction as follow:
If $a = b$ then they just pass by.
Prove that, there is a scheme when people can go from one side to the other side of the road.
Post Script: This is yet a quite interesting problem to solve, especially when you already have a classic original question and you want to elevate it.
In terms of the ant question, it would not be possible for all ants to end up on the opposite side from which they started. Because the ants always reverse direction when they bump into one another, the first ant from the left direction and the first ant from the right direction will never be able to pass one another. It may be possible for all ants from one side to end up on the other side though, if you are allowed to adjust the velocities of the ants.