Solution: The final losing position is a single stone, and the player who creates this state wins. Note that 1 is an odd number, which is a significant characteristic of a losing position. Therefore, we call a state with an odd number of stones an odd state. If a player can make a clever move to always reach a losing state:

Naturally, for any odd state, after one operation, only an odd number of positive divisors of the stones can be taken out, resulting in an even state. For an even state, the player can take away one stone, leaving an odd number of stones, thus reaching an odd state.

In summary, when $n$ is odd, the given state is a losing position, meaning the second player has a winning strategy; when $n$ is even, the given state is a winning position, meaning the first player has a winning strategy.

Using a similar method, we can solve a problem from the 1989 Leningrad Mathematical Olympiad:

Two people play a game, with the number 2 written on the blackboard. They take turns replacing the number $n$ on the blackboard with $n+d$, where $d$ is any positive divisor of $n$ other than itself. The player who writes a number greater than 19891989 loses. Who has the winning strategy?