Four, (50 points) Distribute 2010 red cards and 2010 white cards arbitrarily to 2010 players participating in the game, with each person receiving two cards. All players sit in a circle facing down. The game rule is that each operation requires each player to simultaneously follow the following principle: if a player has at least one red card, he will give a red card to the player on his left; if he has no red card, he will give a white card to the player on his left. Find the maximum number of operations required for the first time that each player has exactly one red card and one white card in their hands.
In 2009.
Let $n=2$ 010. A wheel (divided into $n$ sections) can be placed in a circle of $n$ players. Each player, according to the game rules, when the section they face contains a white card and they hold a red card, they place the red card in the section and take back the white card; in all other cases, it is assumed that no card exchange is necessary.

The 0th operation is set as all players placing a card in the section they face (players with a red card place a red card, otherwise, they place a white card), then the wheel rotates one section clockwise.

The following proof shows: the wheel will have all red cards after at most $n-1$ rotations, meaning each player will have exactly one red and one white card.

By relativity, the wheel can be considered stationary, and the $n$ players move one section counterclockwise around the wheel, with the same card exchange rules. When they move $n-1$ times, each player will have visited all $n$ sections of the wheel, and players with red cards will have placed all of them in a section of the wheel, taking back a white card, making all the cards on the wheel red.

Additionally, if $n$ red cards are given to the $\left[\frac{n+1}{2}\right]$ players adjacent to the left (let's say players 1, 2, ..., $\left[\frac{n+1}{2}\right]$), then initially, the first $\left[\frac{n+1}{2}\right]$ sections of the wheel will have red cards, and the rest will have white cards. Player 1 must place their red card in the $n$th section, meaning the operation has been performed $n-1$ times.
(Song Qiang, compiled)