The first mover, Ivan, has the winning tactic. First, we divide the big rectangle into 503 parts, including a 4x2 ( fault should be 2x2) rectangle on the leftmost region and 502 four times 2 rectangles arranged consecutively from left to right. Ivan is to put his 1st domino in the 4x2 ( fault should be 2x2 )rectangle. Everytime when Peter put his domino in any 4x2 cell, Ivan follows him by putting another donimo in that cell so to make both he and Peter have chance to put 1 extra domino into that cell. If Peter later fill up this cell again, Ivan follows him. It is easy to check that this is possible. In short, Ivan's strategy is to follow Peter in completing any 4x2 cell. As Ivan preserve himself a chance to put 1 more domino in the 4x2 (fault should be 2x2 ) leftmost retangle, after both put equal number of dominos to fill in the 2008x2 rectangle, Ivan wins the game by putting the last horizontal domino on the leftmost.

shing69696 wrote: First, we divide the big rectangle into 503 parts, including a 4x2 rectangle on the leftmost region and 502 four times 2 rectangles arranged consecutively from left to right. There's a little fault: It should be 2x2 square, not 4x2 rectangle.

Thanks for pointing out the error. I have changed already.

shing69696 wrote: The first mover, Ivan, has the winning tactic. First, we divide the big rectangle into 503 parts, including a 4x2 ( fault should be 2x2) rectangle on the leftmost region and 502 four times 2 rectangles arranged consecutively from left to right. Ivan is to put his 1st domino in the 4x2 ( fault should be 2x2 )rectangle. Everytime when Peter put his domino in any 4x2 cell, Ivan follows him by putting another donimo in that cell so to make both he and Peter have chance to put 1 extra domino into that cell. If Peter later fill up this cell again, Ivan follows him. It is easy to check that this is possible. In short, Ivan's strategy is to follow Peter in completing any 4x2 cell. As Ivan preserve himself a chance to put 1 more domino in the 4x2 (fault should be 2x2 ) leftmost retangle, after both put equal number of dominos to fill in the 2008x2 rectangle, Ivan wins the game by putting the last horizontal domino on the leftmost. I don't really understand your solution. Could you say it more clearly (with a figure), please?

Let me rephrase it. Divide the board to 503 parts: a 2x2 square at one end of the board, and the rest are divided into 4x2 rectangles. Ivan put a 2x1 domino in the 2x2 square, securing the other half of the square for him. Now, Peter must put a 1x2 domino in a 4x2 rectangle. Ivan can put a 2x1 domino in the same rectangle, thus ensuring that both Peter and Ivan can put a domino each in the same rectangle again.* This way, the 4x2 rectangles must be completely filled. Ivan has secured a play for himself, thus Peter will be guaranteed to lose. *Proof: If Peter plays either of the red dominoes, Ivan can simply play either of the green dominoes. When Peter plays the other red domino, Ivan simply play the other green domino, thus ensuring that every 4x2 rectangle can be covered. (If Peter plays at the right side of the rectangle, simply flip the image.)

Attachments:

What if the 1X2 domino is placed in between two consecutive 2X4 rectangles ?

The board is horizontal; it's impossible for a vertical 1x2 to span over two columns.

By 1X2, I meant 1 row X 2 columns .. i.e. a horizontal domino... What if this horizontal domino was placed in the middle of two of these 2X4 units (again 2 rows X 4 cols) and not strictly inside any one?

The vertical one is placed first

If Ivan wants to follow that strategy to ensure the win, Ivan will never make such move.