From 1, 2 How many numbers can be taken out of 2010 positive integers, so that the sum of any three numbers can be divided by 33?

From 1, 2 How many numbers can be taken out of 2010 positive integers, so that the sum of any three numbers can be divided by 33?

First of all, the following 61 numbers: 11, 11 + 33, 11 + 2 × 33, 11 + 60 × 33 (1991) satisfy the condition of the problem. On the other hand, let A1 ＜ A2 ＜ an be the number from 1, 22010 that satisfies the condition of the problem. For any four numbers AI, AJ, AK, am in these n numbers, because 33 | (AI + AK + AM), 33 | (AJ + AK + AM), so the difference between any two numbers taken by 33 | (AJ AI),!) is 33 Let AI = a1 + 33di, I = 1, 2, 3, N, from 33 | (a1 + A2 + a3), get 33 | (3A1 + 33d2 + 33d3), so 33 | 3A1, 11 | A1, that is, A1 ≥ 11, DN = an − A133 ≤ 2010 − 1133 < 61, so DN ≤ 60, so n ≤ 61. In conclusion, the maximum value of n is 61