3. 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 any other player, what is the maximum number of players on this team?
3. First, all the elements in the set of all odd numbers $\{1,3, \cdots, 99\}$, which are 50 in total, can be the numbers of the team members. Now we prove: it is impossible to increase further. If the sports team has 51 members, their numbers, in ascending order, are denoted as $a_{1}, a_{2}, \cdots, a_{50}, a_{51}$. (1) Calculate the differences $a_{51}-a_{1}, a_{51}-$ $a_{2}, \cdots, a_{51}-a_{50}$, (2) which gives us $51+50=101$ numbers. Since the numbers in (1) and (2) do not exceed the range of $1 \sim 100$, 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), giving $a_{i}=a_{51}-a_{j}$, which means $a_{i}+a_{j}=a_{51}$. This indicates that the number $a_{51}$ is equal to the sum of the numbers of two other members (when $i \neq j$), or is equal to twice the number of one member (when $i=j$), which contradicts the condition. Therefore, the maximum number of members is 50.