For every positive integer $x$, let $k(x)$ denote the number of composite numbers that do not exceed $x$. 
Find all positive integers $n$ for which  $(k (n))! $ lcm $(1, 2,..., n)> (n - 1) !$ .
To solve the problem, we need to find all positive integers \( n \) for which \((k(n))! \cdot \text{lcm}(1, 2, \ldots, n) > (n-1)!\). Here, \( k(n) \) denotes the number of composite numbers that do not exceed \( n \).

1. **Understanding \( k(n) \)**:
   - \( k(n) \) is the count of composite numbers less than or equal to \( n \).
   - Composite numbers are numbers that have more than two distinct positive divisors.

2. **Rewriting the inequality**:
   - The given inequality is \((k(n))! \cdot \text{lcm}(1, 2, \ldots, n) > (n-1)!\).
   - The least common multiple (LCM) of the first \( n \) natural numbers can be expressed in terms of their prime factorization: \(\text{lcm}(1, 2, \ldots, n) = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\), where \( p_i \) are primes and \( e_i \) are their respective exponents.

3. **Analyzing the inequality**:
   - We need to compare \((k(n))! \cdot p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\) with \((n-1)!\).
   - For large \( n \), \((n-1)!\) grows very quickly, so we need to find \( n \) such that the product of the factorial of the number of composite numbers and the LCM of the first \( n \) numbers exceeds \((n-1)!\).

4. **Considering \( n \) of the form \( p^k \)**:
   - If \( n \) is not of the form \( p^k \) (where \( p \) is a prime and \( k \) is an integer), it is unlikely that the inequality will hold because the right-hand side \((n-1)!\) will be too large.
   - Therefore, we focus on \( n \) of the form \( p^k \).

5. **Checking small values of \( n \)**:
   - We need to check small values of \( n \) to see if they satisfy the inequality.
   - For \( n = 2 \):
     \[
     k(2) = 0 \quad \text{(no composite numbers less than or equal to 2)}
     \]
     \[
     (k(2))! \cdot \text{lcm}(1, 2) = 1 \cdot 2 = 2
     \]
     \[
     (2-1)! = 1
     \]
     \[
     2 > 1 \quad \text{(True)}
     \]
   - For \( n = 3 \):
     \[
     k(3) = 0 \quad \text{(no composite numbers less than or equal to 3)}
     \]
     \[
     (k(3))! \cdot \text{lcm}(1, 2, 3) = 1 \cdot 6 = 6
     \]
     \[
     (3-1)! = 2
     \]
     \[
     6 > 2 \quad \text{(True)}
     \]
   - For \( n = 4 \):
     \[
     k(4) = 1 \quad \text{(4 is the only composite number less than or equal to 4)}
     \]
     \[
     (k(4))! \cdot \text{lcm}(1, 2, 3, 4) = 1 \cdot 12 = 12
     \]
     \[
     (4-1)! = 6
     \]
     \[
     12 > 6 \quad \text{(True)}
     \]
   - For \( n = 5 \):
     \[
     k(5) = 1 \quad \text{(4 is the only composite number less than or equal to 5)}
     \]
     \[
     (k(5))! \cdot \text{lcm}(1, 2, 3, 4, 5) = 1 \cdot 60 = 60
     \]
     \[
     (5-1)! = 24
     \]
     \[
     60 > 24 \quad \text{(True)}
     \]
   - For \( n = 7 \):
     \[
     k(7) = 2 \quad \text{(4 and 6 are the composite numbers less than or equal to 7)}
     \]
     \[
     (k(7))! \cdot \text{lcm}(1, 2, 3, 4, 5, 6, 7) = 2! \cdot 420 = 2 \cdot 420 = 840
     \]
     \[
     (7-1)! = 720
     \]
     \[
     840 > 720 \quad \text{(True)}
     \]
   - For \( n = 9 \):
     \[
     k(9) = 4 \quad \text{(4, 6, 8, and 9 are the composite numbers less than or equal to 9)}
     \]
     \[
     (k(9))! \cdot \text{lcm}(1, 2, 3, 4, 5, 6, 7, 8, 9) = 4! \cdot 2520 = 24 \cdot 2520 = 60480
     \]
     \[
     (9-1)! = 40320
     \]
     \[
     60480 > 40320 \quad \text{(True)}
     \]

6. **Conclusion**:
   - The only values of \( n \) that satisfy the inequality are \( 2, 3, 4, 5, 7, \) and \( 9 \).

The final answer is \( \boxed{ \{2, 3, 4, 5, 7, 9\} } \)