Find all positive integers $n$ that have exactly $16$ positive integral divisors $d_{1},d_{2} \cdots, d_{16}$ such that $1=d_{1}<d_{2}<\cdots<d_{16}=n$, $d_6=18$, and $d_{9}-d_{8}=17$.
To solve the problem, we need to find all positive integers \( n \) that have exactly 16 positive integral divisors such that \( 1 = d_1 < d_2 < \cdots < d_{16} = n \), \( d_6 = 18 \), and \( d_9 - d_8 = 17 \).

1. **Determine the form of \( n \):**
   Since \( n \) has exactly 16 divisors, we can use the divisor function. If \( n \) has the prime factorization \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \), then the number of divisors of \( n \) is given by:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) = 16
   \]
   The possible factorizations of 16 are:
   \[
   16 = 16 \quad (1 \text{ factor})
   \]
   \[
   16 = 8 \cdot 2 \quad (2 \text{ factors})
   \]
   \[
   16 = 4 \cdot 4 \quad (2 \text{ factors})
   \]
   \[
   16 = 4 \cdot 2 \cdot 2 \quad (3 \text{ factors})
   \]
   \[
   16 = 2 \cdot 2 \cdot 2 \cdot 2 \quad (4 \text{ factors})
   \]
   Since \( d_6 = 18 \), we need to consider the prime factorization of 18, which is \( 2 \cdot 3^2 \). This suggests that \( n \) must include these factors.

2. **Form of \( n \) based on \( d_6 = 18 \):**
   Given \( d_6 = 18 \), we can infer that \( n \) must be of the form \( 2 \cdot 3^3 \cdot p \) for some prime \( p \). This is because 18 is the 6th smallest divisor, and the prime factorization of \( n \) must accommodate this.

3. **Condition \( d_9 - d_8 = 17 \):**
   We need to find \( k \) such that \( k(k + 17) = n = 54p \). Considering modulo 27, we get:
   \[
   k(k + 17) \equiv 0 \pmod{27}
   \]
   This implies:
   \[
   k \equiv 0 \text{ or } 10 \pmod{27}
   \]

4. **Case 1: \( k \equiv 0 \pmod{27} \):**
   Let \( k = 27a \). Then:
   \[
   27a(27a + 17) = 54p
   \]
   Simplifying, we get:
   \[
   729a^2 + 459a = 54p
   \]
   \[
   27a(27a + 17) = 54p
   \]
   \[
   a(27a + 17) = 2p
   \]
   Since \( a \) and \( 27a + 17 \) are relatively prime, \( a \) must be 1 or 2 to keep \( p \) prime:
   - If \( a = 1 \):
     \[
     1(27 \cdot 1 + 17) = 2p \implies 44 = 2p \implies p = 22 \quad (\text{not prime})
     \]
   - If \( a = 2 \):
     \[
     2(27 \cdot 2 + 17) = 2p \implies 2 \cdot 71 = 2p \implies p = 71 \quad (\text{prime})
     \]

5. **Case 2: \( k \equiv 10 \pmod{27} \):**
   Let \( k = 27a + 10 \). Then:
   \[
   (27a + 10)(27a + 27) = 54p
   \]
   Simplifying, we get:
   \[
   (a + 1)(27a + 10) = 2p
   \]
   Since \( a + 1 \leq 2 \):
   - If \( a = 1 \):
     \[
     (1 + 1)(27 \cdot 1 + 10) = 2p \implies 2 \cdot 37 = 2p \implies p = 37 \quad (\text{prime})
     \]

Thus, the only possible working values of \( n \) are:
\[
\boxed{1998, 3834}
\]
(54 times 37 and 71 respectively). It can easily be verified that they work, so we're done.