9th grade AMTI question $-$ $65$ bugs on a $9 \times 9$ board 65 bugs are placed at different squares of 9X9 square board. A bug in each moves to a horizontal or vertical adjacent square. No bug makes two horizontal or two vertical moves in succession. Show that after some moves, there will be at least two bugs in the same square.
 A: Jack Frost's link contains a brief solution from v_Enhance at the Art of Problem Solving, but that answer was deleted by the owner. So here is an illustrated version:

There are $16$ red squares, so in four turns, at most $64$ bugs can visit a red square without colliding. But you should be able to convince yourself that each bug must visit a red square on every fourth turn. Hence, if there are no collisions, there can be at most $64$ bugs.
Remark: This argument shows that if there are $65$ bugs, then two bugs will collide after at most three moves; my comment to the OP shows that two moves are not enough.
A: This is a question based on topology of the bugs on the board.
So for that reason the bugs can be placed in any place on the board as the question mentions only that the bugs are placed on the different place on the board.
Now, Please refer the diagram.The Diagram of the board
In the diagram if one observes carefully. the board can be divided into boxes of 4X4.
The reason the boxes are of 4X4 is chosen is because 64+1=65. And 64 is a multiple of 4.
When the boxes are arranged in the 9X9 board as shown a corridor of 17 boxes are left in the corner perimeter. This is done to prove the question.
Now in all 4X4 boxes shown by mono color there are 4 bugs. These bugs will move in either clockwise or counter clockwise motion. 


*

*Just to give you an example. The Upper Left bug will come to the lower left, the lower left bug will move to the lower right, the lower right bug will go the upper right. and cycle will continue. This sort of ordered motion is done to symplify the problem

*The bugs will never leave their individual boxes. This will not violate the rule of alternate successive motion.

*Now we do some simple math to find out that only one bug is left. This bug can be placed in any place in the outer corridor. It is now easy to visualize that bug may move the first time. But the next time it cannot go to any empty box. (no horizontal motion if fist move is vertical or vice versa )

*Therefore after some moves there will have to be at least 2 bugs in a block. In this case just after one move

*Solution by Mechanical Engineering Student.

