We call a divisor $d$ of a positive integer $n$ [i]special[/i] if $d + 1$ is also a divisor of $n$. Prove: at most half the positive divisors of a positive integer can be special. Determine all positive integers for which exactly half the positive divisors are special.
1. **Prove that at most half the positive divisors of a positive integer \( n \) can be special:**

   Let \( d \) be a special divisor of \( n \). This means \( d \mid n \) and \( d+1 \mid n \). Consider the case where \( d \geq \sqrt{n} \). Since \( \gcd(d, d+1) = 1 \), if both \( d \) and \( d+1 \) are divisors of \( n \), then \( d(d+1) \mid n \). However, \( d(d+1) > d^2 \geq n \), which is a contradiction. Therefore, no divisor \( d \geq \sqrt{n} \) can be special.

   Since the divisors of \( n \) come in pairs \( (d, \frac{n}{d}) \), and for each \( d < \sqrt{n} \), \( \frac{n}{d} > \sqrt{n} \), at most half of the divisors of \( n \) are less than \( \sqrt{n} \). Hence, at most half of the divisors of \( n \) can be special.

2. **Determine all positive integers for which exactly half the positive divisors are special:**

   If exactly half of the positive divisors of \( n \) are special, then \( n \) cannot be a perfect square because \( \sqrt{n} \) would be a divisor, and it cannot be special as shown above. For each divisor \( d \) of \( n \) with \( d < \sqrt{n} \), \( d+1 \) must also be a divisor of \( n \).

   Since \( 1 \mid n \), every positive integer \( k \leq \lceil \sqrt{n} \rceil \) must divide \( n \). This implies that \( n \) must be highly composite. Specifically, \( n \) must be even. If \( n \geq 5 \), then \( n \) must also be divisible by 3, and if \( n \geq 17 \), also by 5.

   For \( n \leq 25 \), we check the numbers:
   - \( n = 2 \): Divisors are \( 1, 2 \). Special divisors: \( 1 \) (since \( 2 \mid 2 \)). Exactly half are special.
   - \( n = 6 \): Divisors are \( 1, 2, 3, 6 \). Special divisors: \( 1, 2 \) (since \( 2 \mid 6 \) and \( 3 \mid 6 \)). Exactly half are special.
   - \( n = 12 \): Divisors are \( 1, 2, 3, 4, 6, 12 \). Special divisors: \( 1, 2, 3 \) (since \( 2 \mid 12 \), \( 3 \mid 12 \), and \( 4 \mid 12 \)). Exactly half are special.

   For \( n > 25 \), let \( r = \lfloor \sqrt{n} \rfloor \). Then \( r-1, r, r+1 \) must divide \( n \), and the greatest common divisor of any two of them is 2. Thus,
   \[
   \frac{(r-1)r(r+1)}{2} \mid n \implies (r-1)r(r+1) \leq 2n \implies \sqrt{n}(\sqrt{n}-1)(\sqrt{n}-2) < 2n \implies n+2 < 5\sqrt{n}.
   \]
   This inequality does not hold for \( n > 25 \). Therefore, the only possibilities are \( n = 2, 6, 12 \).

The final answer is \( \boxed{ n = 2, 6, 12 } \).