Find all positive integers $ n$ having exactly $ 16$ divisors $ 1\equal{}d_1<d_2<...<d_{16}\equal{}n$ such that $ d_6\equal{}18$ and $ d_9\minus{}d_8\equal{}17.$
To find all positive integers \( n \) having exactly 16 divisors such that \( d_6 = 18 \) and \( d_9 - d_8 = 17 \), we start by analyzing the number of divisors.

1. **Determine the form of \( n \):**
   If \( n = \prod p_i^{n_i} \), the number of divisors of \( n \) is given by:
   \[
   \prod (n_i + 1) = 16
   \]
   The possible factorizations of 16 are:
   \[
   16 = 2 \times 2 \times 2 \times 2 \quad \Rightarrow \quad n = p_1 p_2 p_3 p_4
   \]
   \[
   16 = 2 \times 2 \times 4 \quad \Rightarrow \quad n = p_1 p_2 p_3^3
   \]
   \[
   16 = 4 \times 4 \quad \Rightarrow \quad n = p_1^3 p_2^3
   \]
   \[
   16 = 2 \times 8 \quad \Rightarrow \quad n = p_1 p_2^7
   \]
   \[
   16 = 16 \quad \Rightarrow \quad n = p_1^{15}
   \]

2. **Analyze the given conditions:**
   Since \( d_6 = 18 = 2 \times 3^2 \), we can eliminate the forms \( n = p_1 p_2 p_3 p_4 \) and \( n = p_1^{15} \) because they cannot produce 18 as the 6th smallest divisor. We are left with:
   \[
   n = 2 \times 3^3 \times p \quad \text{(with prime } p > 3)
   \]
   \[
   n = 2^3 \times 3^3
   \]
   \[
   n = 2 \times 3^7
   \]

3. **Check each remaining form:**

   - **Case 1: \( n = 2 \times 3^3 \times p \)**
     The divisors are:
     \[
     1, 2, 3, 6, 9, 18, 27, 54, p, 2p, 3p, 6p, 9p, 18p, 27p, 54p
     \]
     Since \( d_6 = 18 \), \( p > 18 \). We need \( d_9 - d_8 = 17 \).

     - **Subcase 1.1: \( 18 < p < 27 \)**
       Here, \( d_8 = 27 \) and \( d_9 = 2p \). Thus:
       \[
       2p - 27 = 17 \quad \Rightarrow \quad 2p = 44 \quad \Rightarrow \quad p = 22 \quad \text{(not prime)}
       \]

     - **Subcase 1.2: \( 27 < p < 54 \)**
       Here, \( d_8 = p \) and \( d_9 = 54 \). Thus:
       \[
       54 - p = 17 \quad \Rightarrow \quad p = 37 \quad \text{(prime)}
       \]
       Therefore, \( n = 2 \times 3^3 \times 37 = 1998 \).

     - **Subcase 1.3: \( 54 < p \)**
       Here, \( d_8 = 54 \) and \( d_9 = p \). Thus:
       \[
       p - 54 = 17 \quad \Rightarrow \quad p = 71 \quad \text{(prime)}
       \]
       Therefore, \( n = 2 \times 3^3 \times 71 = 3834 \).

   - **Case 2: \( n = 2^3 \times 3^3 \)**
     The divisors are:
     \[
     1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216
     \]
     Here, \( d_6 = 8 \), which does not satisfy \( d_6 = 18 \). Thus, this case is invalid.

   - **Case 3: \( n = 2 \times 3^7 \)**
     The divisors are:
     \[
     1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374
     \]
     Here, \( d_6 = 18 \), but \( d_9 - d_8 = 81 - 54 = 27 \), which does not satisfy \( d_9 - d_8 = 17 \). Thus, this case is invalid.

Conclusion:
The only valid solutions are \( n = 1998 \) and \( n = 3834 \).

The final answer is \( \boxed{1998, 3834} \)