First, we can take the numbers 1 1, 33+1,33x2+ 1 1, 33x60+ 1 1, so that the sum of every three.

Secondly, it is proved that 62 cannot take any number, and every sum of three can be divisible by 33. (2)

Proved as follows:

First, divide 1 ~ 20 10 into 33 groups according to the remainder divisible by 33, namely

1 group: 1, 33+ 1, 33× 2+ 1, ..., 33× 60+ 1.

This group has 6 1 numbers.

The second group: 2,33+2,33× 2+2,33× 60+2.

This group has 6 1 numbers.

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Group 33: 33, 33x2, 33x3, ... 33x60.

There are 60 numbers in this group.

Each group has a maximum of 6 1 and a minimum of 60.

If there are 62 numbers, the sum of every three numbers can be divisible by 33;

Then, because the sum of every three numbers can be divisible by 33, when any two numbers are taken and one of the remaining 60 numbers is taken, because any one of these 60 numbers can be divisible by 33, that is, their remainder divisible by 33 should be the same, so these 60 numbers should be in the same group; Similarly, if two of the 60 numbers are taken, and the other 60 numbers in the 62 numbers are also in the same group, then the 62 numbers must all be in the same group. This is in contradiction with the previous "6 1 number per group at most and 60 numbers per group at least". So it is impossible to take any number of 62, and the sum of every three can be divisible by 33.

Combining (1) and (2), it is found that there can be at most 6 1 number, so that the sum of any three numbers can be divisible by 33.