1+2+。 . . +2009=2009*20 10/2= 1050*2009

= 999 * 2009+6 * 2009 = 999 * 2009+6 * 999+6 *11.Obviously, this value is divided by 9+3.

It is also proved that the number N= 123...2009 and the remainder divided by 9 equals to M= 1+2+3+...+2009 and divided by 9. This proposition should be correct, but it is not easy to prove.

Method 2

Analysis:

1, from 0 to 9, and the sum is 45, and the top ten digits can be divisible by 9.

2. From 10 to 19, because all the digits add up, compared with the above, there are only 10/more.

3. From 0 to 99, from the analysis of 2, it can be seen that only the addition of ten digits is equivalent to the addition of 0 to 9, and10 * (1+2+...+9) = 450. So the first hundred digits can be divisible by 9.

4. From 0 to 999, 3 is only the addition of hundreds of digits,100 * (1+2+...+9) = 4500, and the first 1000 digits can be divisible by 9.

5. From 0 to 1999, which is equivalent to from 1000 to 1999, only thousands of digits are added, which is 1000. Divided by 9, it is 1.

6. From 0 to 2009, it is equivalent to the sum of 2000 to 2009 plus 1. However, from 2000 to 2009, only thousands of digits added up to 20. Add the remainder of the first 2000 numbers 1 and divide it by 9, which is 2 1. If you divide it by 9, the remainder is 3.