Let B={ 1, 3, 5, …, 2n+ 1}, then the sum of any two numbers in this set is even, which accords with the meaning of the question.

It is proved that the sum of two numbers must still be in the set B composed of any n+2 elements in A. 。

Prove by mathematical induction.

When n= 1, A={ 1, 2,3}, let the set of three elements in a be B={ 1, 2,3}, which obviously has 1+2=3, and the conclusion is valid.

Assuming n, the conclusion holds, that is, the sum of two numbers in the set b of any n+2 elements in a = {1, 2, 3, …, 2n, 2n+ 1} must still be in b. 。

For n+ 1, a = {1, 2, 3, …, 2n+ 1, 2n+2, 2n+3}, choose any n+3 elements from A to form the set B, and prove that the sum of two numbers in B is still in B. 。

If the number of n+3 does not contain 2n+2 or 2n+3, then the number of n+2 must be taken from {1, 2,3, ..., 2n, 2n+ 1}. By inductive hypothesis, the sum of two numbers must be in B, and the conclusion is valid.

If the number of n+3 includes 2n+2 and 2n+3, the number of n+ 1, 2,3, …, 2n, 2n+ 1 should be taken out. It is proved that 2n+3 must be expressed as the sum of two numbers in b 。

Divide the 2n+2 numbers of 1, 2, 3, …, 2n+ 1, 2n+2 into n+ 1 groups (1, 2n+2), (2, 2n+ 1), (. There must be two numbers of n+2 in the same group. Because 2n+3 =1+(2n+2) = 2+(2n+1) = 3+2n = … = (n+1)+(n+2), it is at 65438.

According to the principle of mathematical induction, it is known that the sum of two numbers in set B, which is composed of n+2 numbers in set A, must still be in B. 。

Therefore, the maximum number of elements in B is n+ 1.