The first round: remainder 2, 4, 6, 8... 10002n.

Round 2: Remaining 4,8, 12, 16 1000 4n.

Round 3: Remaining 816,2410008n.

M-th round: m power n of remainder 2

Round 9: Remaining 5 12

The last one is 5 12, which is the maximum value that can be taken by the m power of 2 in 1000.

2. From the extreme observation, we assume that B is the player who won the most times, but because B didn't win all the games, there must be a player who won B. Among the players who lost to B, there must be a player who won A. Otherwise, A won more times than B, which contradicts B's winning the most times. So we must find three players, A, B and C, so that A wins B, B wins C and C wins A. ..

3.

Using logical reasoning, it is assumed that the difference between the numbers filled in any two adjacent squares is less than or equal to 4. Then consider the location of 1. If 1 is in the middle, then it has four neighbors, only 2, 3, 4, 5. Except these four neighbors, 2 can only be 6, which means 2 can only have 1 and 6 neighbors, so 2 can only be in the four corners. But 1 is in the middle, and neighbor 2 of 1 can't be in the four corners. Judging from this, 1 can't be in the middle, only nearby. Similarly, it can be inferred that 1 can't be on four corners.

Let's examine the parity of the sum of all the numbers in the table: before the first operation, it is equal to 9, which is an odd number. Every time we "operate", we have to change the parity of four squares in a row or a column. Obviously, the parity of the sum of all numbers in the whole 16 grid remains unchanged. But when all the numbers in each cell become 1, the sum of all the numbers in the whole 16 cell is 16, which is an even number. So it is impossible to change the number in each cell into 1 through several operations.