Find all positive integers $k$ such that the product of the digits of $k$, in decimal notation, equals \[\frac{25}{8}k-211\]
1. **Identify the condition for \( k \):**
   The problem states that the product of the digits of \( k \) equals \(\frac{25}{8}k - 211\). Since the product of the digits of \( k \) is an integer, \(\frac{25}{8}k - 211\) must also be an integer. This implies that \( \frac{25}{8}k \) must be an integer, so \( k \) must be a multiple of 8. Let \( k = 8n \) for some integer \( n \).

2. **Rewrite the equation:**
   Substitute \( k = 8n \) into the given equation:
   \[
   \text{Product of the digits of } k = \frac{25}{8}(8n) - 211 = 25n - 211
   \]

3. **Determine the parity of \( n \):**
   Since \( k \) is a multiple of 8, the rightmost digit of \( k \) must be even. Therefore, the product of the digits of \( k \) (which is \( 25n - 211 \)) must also be even. For \( 25n - 211 \) to be even, \( 25n \) must be odd (since subtracting an odd number, 211, from an odd number results in an even number). Therefore, \( n \) must be odd.

4. **Establish bounds for \( n \):**
   The product of the digits of \( k \) is at most \( k \). Therefore:
   \[
   25n - 211 \leq 8n
   \]
   Simplify this inequality:
   \[
   25n - 8n \leq 211 \implies 17n \leq 211 \implies n \leq \frac{211}{17} \approx 12.41
   \]
   Since \( n \) is an integer, \( n \leq 12 \).

   Additionally, the product of the digits of \( k \) must be non-negative:
   \[
   25n - 211 \geq 0 \implies 25n \geq 211 \implies n \geq \frac{211}{25} \approx 8.44
   \]
   Since \( n \) is an integer, \( n \geq 9 \).

5. **Check possible values of \( n \):**
   Since \( n \) must be an odd integer between 9 and 12, the possible values for \( n \) are 9 and 11.

   - For \( n = 9 \):
     \[
     k = 8 \times 9 = 72
     \]
     The product of the digits of 72 is \( 7 \times 2 = 14 \).
     Check the equation:
     \[
     25 \times 9 - 211 = 225 - 211 = 14
     \]
     This works.

   - For \( n = 11 \):
     \[
     k = 8 \times 11 = 88
     \]
     The product of the digits of 88 is \( 8 \times 8 = 64 \).
     Check the equation:
     \[
     25 \times 11 - 211 = 275 - 211 = 64
     \]
     This works.

Conclusion:
Both \( k = 72 \) and \( k = 88 \) satisfy the given condition.

The final answer is \( \boxed{72 \text{ and } 88} \)