For $n$ a positive integer, denote by $P(n)$ the product of all positive integers divisors of $n$. Find the smallest $n$ for which 

\[ P(P(P(n))) > 10^{12} \]
1. **Understanding the formula for \( P(n) \):**
   The product of all positive divisors of \( n \) is given by:
   \[
   P(n) = n^{\tau(n)/2}
   \]
   where \( \tau(n) \) is the number of positive divisors of \( n \).

2. **Analyzing the formula for prime numbers:**
   If \( n = p \) where \( p \) is a prime number, then \( \tau(p) = 2 \) (since the divisors are \( 1 \) and \( p \)). Thus:
   \[
   P(p) = p^{2/2} = p
   \]
   Therefore, \( P(P(P(p))) = p \), which does not satisfy \( P(P(P(n))) > 10^{12} \).

3. **Analyzing the formula for \( n = p^a \):**
   If \( n = p^a \) where \( p \) is a prime number and \( a \) is a positive integer, then \( \tau(p^a) = a + 1 \). Thus:
   \[
   P(p^a) = (p^a)^{(a+1)/2} = p^{a(a+1)/2}
   \]
   We need to find the smallest \( n \) such that \( P(P(P(n))) > 10^{12} \).

4. **Testing \( n = 6 \):**
   Let's test \( n = 6 \):
   - The divisors of \( 6 \) are \( 1, 2, 3, 6 \), so \( \tau(6) = 4 \).
   - Therefore:
     \[
     P(6) = 6^{4/2} = 6^2 = 36
     \]
   - Next, we calculate \( P(36) \):
     - The divisors of \( 36 \) are \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \), so \( \tau(36) = 9 \).
     - Therefore:
       \[
       P(36) = 36^{9/2} = 36^{4.5} = 6^9 = 2^9 \cdot 3^9 = 512 \cdot 19683 = 10077696
       \]
   - Finally, we calculate \( P(10077696) \):
     - The number \( 10077696 \) is \( 2^{18} \cdot 3^9 \).
     - The number of divisors \( \tau(10077696) = (18+1)(9+1) = 19 \cdot 10 = 190 \).
     - Therefore:
       \[
       P(10077696) = (10077696)^{190/2} = (2^{18} \cdot 3^9)^{95} = 2^{1710} \cdot 3^{855}
       \]
     - We need to check if this value is greater than \( 10^{12} \):
       \[
       2^{1710} \cdot 3^{855} > 10^{12}
       \]
       Since \( 2^{10} \approx 10^3 \) and \( 3^5 \approx 10^2 \), we can approximate:
       \[
       2^{1710} \approx (10^3)^{171} = 10^{513}
       \]
       \[
       3^{855} \approx (10^2)^{171} = 10^{342}
       \]
       \[
       2^{1710} \cdot 3^{855} \approx 10^{513} \cdot 10^{342} = 10^{855}
       \]
       Clearly, \( 10^{855} > 10^{12} \).

Therefore, \( n = 6 \) satisfies \( P(P(P(n))) > 10^{12} \).

The final answer is \( \boxed{6} \).