Positive integers $a,b,c$ satisfy $b\leq c$ and
$$\frac{(ab-1)(ac-1)}{bc}=2023.$$
Find all positive values of $c$. 
1. Given the equation:
   \[
   \frac{(ab-1)(ac-1)}{bc} = 2023
   \]
   We start by simplifying the equation. Multiply both sides by \(bc\):
   \[
   (ab-1)(ac-1) = 2023bc
   \]

2. Expand the left-hand side:
   \[
   ab \cdot ac - ab - ac + 1 = 2023bc
   \]
   Simplify:
   \[
   a^2bc - ab - ac + 1 = 2023bc
   \]

3. Rearrange the equation:
   \[
   a^2bc - 2023bc = ab + ac - 1
   \]
   Factor out \(bc\) on the left-hand side:
   \[
   bc(a^2 - 2023) = ab + ac - 1
   \]

4. Since \(b \leq c\), we need to find the possible values of \(a\). We know:
   \[
   a^2 > 2023 \implies a \geq \lceil \sqrt{2023} \rceil = 45
   \]
   and
   \[
   (a-1)^2 \leq 2023 \implies a \leq \lfloor \sqrt{2023} \rfloor + 1 = 45
   \]
   Therefore, \(a = 45\).

5. Substitute \(a = 45\) into the equation:
   \[
   (45b-1)(45c-1) = 2023bc
   \]

6. Expand and simplify:
   \[
   2025bc - 45b - 45c + 1 = 2023bc
   \]
   Rearrange:
   \[
   2bc - 45b - 45c + 1 = 0
   \]

7. Rearrange to factor:
   \[
   (2b - 45)(2c - 45) = 2023
   \]

8. Since \(2023 = 7 \times 17^2\), the factors of \(2023\) are \(1, 7, 17, 119, 289, 2023\). We need to find pairs \((2b-45, 2c-45)\) such that their product is \(2023\).

9. Check each pair:
   - \(2b - 45 = 1\) and \(2c - 45 = 2023\):
     \[
     2b = 46 \implies b = 23
     \]
     \[
     2c = 2068 \implies c = 1034
     \]
   - \(2b - 45 = 7\) and \(2c - 45 = 289\):
     \[
     2b = 52 \implies b = 26
     \]
     \[
     2c = 334 \implies c = 167
     \]
   - \(2b - 45 = 17\) and \(2c - 45 = 119\):
     \[
     2b = 62 \implies b = 31
     \]
     \[
     2c = 164 \implies c = 82
     \]

10. Since \(b \leq c\), the valid pairs are:
    \[
    (b, c) = (23, 1034), (26, 167), (31, 82)
    \]

The final answer is \(\boxed{82, 167, 1034}\).