A positive proper divisor is a positive divisor of a number, excluding itself. For positive integers $n \ge 2$, let $f(n)$ denote the number that is one more than the largest proper divisor of $n$. Determine all positive integers $n$ such that $f(f(n)) = 2$.
1. We start with the given condition \( f(f(n)) = 2 \). This implies that \( f(n) \) must be a prime number because the only number whose largest proper divisor plus one equals 2 is 1, and the only prime number whose largest proper divisor is 1 is 2. Therefore, \( f(n) \) must be 2.

2. Let \( f(n) = p \), where \( p \) is a prime number. If \( p = 2 \), then the largest proper divisor of \( n \) is 1, meaning that \( n \) itself must be a prime number. This is because the only number whose largest proper divisor is 1 is a prime number.

3. Now, consider the case where \( p > 2 \). Suppose \( n = p_1^{a_1} p_2^{a_2} \cdots p_x^{a_x} \), where \( p_1 < p_2 < \cdots < p_x \) are distinct prime numbers and \( a_1, a_2, \ldots, a_x \) are positive integers. Since \( p - 1 \) is the greatest proper divisor of \( n \), we must have:
   \[
   p - 1 = \frac{n}{p_1} = p_1^{a_1 - 1} p_2^{a_2} \cdots p_x^{a_x}
   \]
   Denote the number \( p_2^{a_2} \cdots p_x^{a_x} \) by \( y \). If \( a_1 > 0 \), since \( p - 1 \) is an even number and \( y \) is an odd number, it follows that \( p_1 = 2 \). Therefore, in this case, \( n = 2(p - 1) \).

4. If \( a_1 = 1 \), then \( p - 1 = y \), but the number \( p - 1 \) is even and the number \( y \) is odd, which is a contradiction. Therefore, \( a_1 \) cannot be 1.

5. In conclusion, \( n \) must either be a prime number or \( n = 2(q - 1) \) where \( q \) is a prime number greater than 2.

The final answer is \( \boxed{ n } \) is a prime number or \( n = 2(q - 1) \) where \( q \) is a prime number greater than 2.