6. On the blackboard, there are two positive integers, one is 2002 and the other is a number less than 2002. If the arithmetic mean of the two numbers is an integer $m$, then the following operation can be performed: one of the numbers is erased and replaced with $m$. The maximum number of times such an operation can be performed is ( ).
Answer: 10
Explanation: Grasp the invariant
Argument, if the sum of two numbers is even, the operation can continue. We can set the difference between the two numbers as a, then the smaller number is (2002-a). After one operation, it will become (2002+2002-a) ÷ 2 = 2002 - a ÷ 2. Each operation reduces the new number on the blackboard by half of the difference. To maximize the number of operations, the difference should be a power of 2, the maximum being 2 to the 10th power, 1024. Thus, the first reduction will be 512, the second 256... up to a maximum of 10 times.
Construction: $m=2002-1024=978 ; 10$ numbers are:
$1490,1234,1106,1042,1010,994,986,982,980,979$