Take any number of n+1 from1to 2N, and at least one number can be divisible by another.

For this new clear question, my thinking is:

[1, 2n] can be divided into two groups: odd set n [1, 3, …, 2N- 1] and even set n [2, 4, …, 2N].

For this problem, in order to get the most numbers, a better strategy is to get

Numbers other than 1 in odd set.

The prime numbers except 2 are all odd numbers, so this number set contains the most prime numbers. But odd numbers are not all prime numbers.

So in [3, ..., 2n- 1] (all odd numbers are prime first. ) They are compatible with each other.

At this time, we must take two more numbers in the even set, which can be proved. No matter how even numbers are taken, either these two even numbers are not coprime, or at least one of these two even numbers is not coprime with an odd number:

Because these two even numbers are obtained, there are only three situations:

1 is a power of 2. 2 p/2 q = 2 (p-q), even numbers are not coprime.

Neither is the power of 2.2. P * 2 a and q * 2 b must not be coprime with odd numbers p and Q.

3. There is a power that is not 2. The number 2 p, q * 2 b must be non-coprime with the obtained odd Q.

In order to get two satisfactory even numbers, in case 3, we must adjust the odd set, reduce an odd Q and get an even number accordingly. Moreover, this even number cannot be a power of 2 (case 1), and odd numbers other than q are non-prime numbers (case 3). Therefore, it can never be adjusted.

According to the above analysis, it is impossible to reduce any other odd numbers and increase the corresponding even numbers on the basis of the optimal access strategy. But the optimal fetching strategy can only take n (N- 1 odd, 1 even) nonreciprocal numbers. So according to pigeonhole principle, if you take another number, it will not be coprime with one of the n numbers.