3. Among the 2007 positive integers $1,2, \cdots, 2007$, what is the maximum number of integers that can be selected such that each selected number is coprime with 2007, and the sum of any three selected numbers is not a multiple of 7?
3. When $1,2, \cdots, 2007$ are divided by 7, the remainders 1, 2, 3, 4, 5 each have $286+1=287$ numbers; the remainders 6, 0 each have 286 numbers.

Among $1,2, \cdots, 2007$, the numbers that are not coprime with 2007 are $3,2 \times 3,3 \times 3, \cdots, 669 \times 3$ and $223,2 \times 223,4 \times 223,5 \times 223,7 \times 223,8 \times 223$.

When these numbers that are not coprime with 2007 are divided by 7, the remainders are $3,6,2,5,1,4,0,3,6,2,5,1,4,0$, $\cdots, 3,6,2,5$ and $6,5,3,2,0,6$.

Thus, among these numbers that are not coprime with 2007, the remainders 1, 2, 3, 4, 5, 6, 0 have $95,97,97,95,97,98,96$ numbers respectively.

Among $1,2, \cdots, 2007$ and coprime with 2007, the remainders 1, 2, 3, 4, 5, 6, 0 have 192, 190, 190, 192, 190, 188, 190 numbers respectively.

To ensure that the sum of any three of the selected numbers is not a multiple of 7, at most 2 numbers with a remainder of 0 can be taken. Since the sum of the remainders (1, 3, 3), (3, 2, 2), (2, 6, 6), (6, 4, 4), (4, 5, 5), (5, 1, 1) and (1, 2, 4), (3, 6, 5) are all multiples of 7, at most 2 groups of the other remainders that are not adjacent in Figure 2 can be taken.

It has been verified that taking all the numbers with remainders 1, 4 from 2 groups, and 2 numbers with a remainder of 0, meets the requirements of the problem, and the number of selected numbers reaches the maximum. Therefore, the maximum number of numbers that can be selected, each of which is coprime with 2007, and the sum of any three of which is not a multiple of 7, is $192+192+2=386$.