$d(n)$ shows the number of positive integer divisors of positive integer $n$. For which positive integers $n$ one cannot find a positive integer $k$ such that $\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )$ is a perfect square.
To solve the problem, we need to determine for which positive integers \( n \) one cannot find a positive integer \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square. 

1. **Understanding the function \( d(n) \)**:
   - The function \( d(n) \) represents the number of positive divisors of \( n \).
   - For example, \( d(6) = 4 \) because the divisors of 6 are 1, 2, 3, and 6.

2. **Initial observations**:
   - If \( n \) is a prime number \( p \), then \( d(p) = 2 \) because the only divisors are 1 and \( p \).
   - If \( n \) is a power of a prime, say \( n = p^k \), then \( d(p^k) = k + 1 \).

3. **Key insight**:
   - We need to find \( n \) such that no matter how many times we apply \( d \), we never get a perfect square.
   - A perfect square is an integer that can be expressed as \( m^2 \) for some integer \( m \).

4. **Inductive step**:
   - Suppose for \( n = 1, 2, \ldots, m-1 \), we can find a \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square.
   - We need to prove this for \( n = m \).

5. **Case analysis**:
   - If \( d(m) \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers, then we need to check if \( d(m) \) can be a perfect square.
   - If \( q > 2 \), then \( q-1 \) is even, and \( p^{q-1} \) is a perfect square.
   - If \( q = 2 \), then \( d(m) = p \), and we need to check if \( m \) can be written as \( r_1^{p-1} \) where \( r_1 \) and \( p \) are primes.

6. **Conclusion**:
   - If \( n \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers, then \( d(n) = q \), \( d(q) = 2 \), and \( d(2) = 2 \), which is not a perfect square.
   - Therefore, for such \( n \), we cannot find a \( k \) such that \(\underbrace{d(\dots d(d}_{k\ \text{times}} (n) \dots )\) is a perfect square.

The final answer is \( \boxed{ n } \) can be represented as \( p^{q-1} \) where \( p \) and \( q \) are prime numbers.