II. (50 points) The player numbers of a sports team are uniquely selected from the positive integers 1 to 100. If the number of any player is neither the sum of the numbers of any other two players nor twice the number of another player, what is the maximum number of players this sports team can have?
Second, all odd numbers $1, 3, \cdots, 99$ totaling 50 can be the numbers of the team members. Below is the proof that it is impossible to increase further. If the sports team

has 51 members, from smallest to largest, denoted as
$$
a_{1}, a_{2}, \cdots, a_{50}, a_{51} \text {. }
$$

Take the differences
$$
a_{51}-a_{1}, a_{51}-a_{2}, \cdots, a_{51}-a_{50},
$$

this results in $51+50=101$ numbers. Since the numbers in (1) and (2) do not exceed the range of 1 to 100, by the pigeonhole principle, there must be two numbers that are equal. However, the numbers in (1) are all distinct, and the numbers in (2) are also all distinct, so it can only be that a number $a_{i}$ in (1) equals a number $a_{51}-a_{j}$ in (2). This gives
$$
a_{i}=a_{51}-a_{j},
$$

thus $a_{i}+a_{j}=a_{51}$.
This means that the number $a_{51}$ is equal to the sum of two other numbers (when $i \neq j$) or equal to twice a certain member's number (when $i=j$), which contradicts the condition. Therefore, the maximum number of members is 50.