Additional questions: write 2000 numbers on the blackboard: 1, 2, 3 , 2000, it is allowed to erase two numbers a and B (a ≥ b) each time and write the three numbers A-B, AB and ab (that is, AB writes twice). After 8000 times of doing so, 10000 numbers are obtained. Q: can these 10000 numbers all be less than 500?

Additional questions: write 2000 numbers on the blackboard: 1, 2, 3 , 2000, it is allowed to erase two numbers a and B (a ≥ b) each time and write the three numbers A-B, AB and ab (that is, AB writes twice). After 8000 times of doing so, 10000 numbers are obtained. Q: can these 10000 numbers all be less than 500?

Since A2 + B2 = (a − b) 2 + (AB) 2 + (AB) 2, the sum of squares of all numbers on the blackboard is always the same +20002=16×2000×2001×4001 ＞16×2000×2000×4000＝2.666… × 109＞2.5×109=5002...

Someone wrote 1, 2, 3.2010 numbers on the blackboard. Now do the following: erase 3 of them and write them
After 1004 times of doing this, there are only two digits left, one is 16, and the other is what?

Because only one digit is left, and 16 is two digits, then the rest must be one digit
Remove 16 from 1-2010, and the remaining sum is 2021039,
If the single digit is 9, the rest must be 9

Someone wrote 1, 2, 3. 2010 numbers on the blackboard
Someone wrote 1, 2, 3.2010 numbers on the blackboard. Now do the following operation: erase three of them at will, and then write the sum of the three numbers. After doing this for 1004 times, there are only two numbers left on the blackboard, one is 16, and what is the other number?

The answer is wrong. It should be 16, not 19
So the answer should be 9:
Calculate the sum of 1 ~ 2010, minus 16, the end number of the remaining part is calculated~
2010 and (1 + 2010) * 2010 / 2 = 2021055
Minus 16 is 2021039
So the remaining number is nine