7. There are a total of 10040 numbers written on the blackboard, including 2006 ones, 2007 twos, 2008 threes, 2009 fours, and 2010 fives. Each operation involves erasing 4 different numbers and writing the fifth type of number (for example, erasing 1, 2, 3, and 4 each once and writing one 5; or erasing 2, 3, 4, and 5 each once and writing one 1, etc.). If after a finite number of operations, there are exactly two numbers left on the blackboard, what is the product of these two numbers?
【Solution】Solution: Since after each operation, the parity of the number of each digit changes, 
what was originally an odd number becomes an even number, and what was originally an even number becomes an odd number.
That is, no matter how the operation is performed, the parity of the original 2 odd numbers remains the same, and the parity of the original 3 even numbers also remains the same.
Finally, 2 numbers $(1, 1)$ are left, indicating that the remaining numbers have the same parity as the original number of digits, so it can only be the original 2 odd numbers (2007, 2009), 2 and 4. $2 \times 4=8$, the result is 8.