Find all positive integers $k$, for which the product of some consecutive $k$ positive integers ends with $k$. 

[i]Proposed by Oleksiy Masalitin[/i]
1. **Analyzing the problem for general \( k \):**
   We need to find all positive integers \( k \) such that the product of some consecutive \( k \) positive integers ends with \( k \). 

2. **Considering the form of \( k \):**
   Let \( k = 10q + r \) where \( q \geq 1 \) and \( 0 \leq r < 10 \). If \( q \geq 2 \), then the product of \( k \) consecutive integers will have at least \( q \) trailing zeros because there will be at least \( q \) multiples of 10 in the product. This means the product will end with at least \( 10q \) zeros, which is more than \( k \) if \( q \geq 2 \). Therefore, \( k \leq 19 \).

3. **Checking \( k \) from 11 to 19:**
   If \( k \) is between 11 and 19, the product of \( k \) consecutive integers will end with a zero because it will include at least one multiple of 10. Hence, \( k \) cannot be between 11 and 19. Thus, \( k \leq 10 \).

4. **Checking \( k \) from 1 to 10:**
   - \( k = 1 \): The product of 1 consecutive integer is just the integer itself, which trivially ends with 1. So, \( k = 1 \) works.
   - \( k = 2 \): Consider the product \( 3 \cdot 4 = 12 \), which ends with 2. So, \( k = 2 \) works.
   - \( k = 3 \): The product of any 3 consecutive integers will be even and thus cannot end with 3. So, \( k = 3 \) does not work.
   - \( k = 4 \): Consider the product \( 1 \cdot 2 \cdot 3 \cdot 4 = 24 \), which ends with 4. So, \( k = 4 \) works.
   - \( k = 5 \): The product of any 5 consecutive integers will be even and a multiple of 5, thus ending in 0. So, \( k = 5 \) does not work.
   - \( k = 6 \): The product of any 6 consecutive integers will be even and a multiple of 5, thus ending in 0. So, \( k = 6 \) does not work.
   - \( k = 7 \): The product of any 7 consecutive integers will be even and a multiple of 5, thus ending in 0. So, \( k = 7 \) does not work.
   - \( k = 8 \): The product of any 8 consecutive integers will be even and a multiple of 5, thus ending in 0. So, \( k = 8 \) does not work.
   - \( k = 9 \): The product of any 9 consecutive integers will be even and a multiple of 5, thus ending in 0. So, \( k = 9 \) does not work.
   - \( k = 10 \): The product of any 10 consecutive integers will include two multiples of 5 and at least five multiples of 2, making it a multiple of 100, thus ending in 00. So, \( k = 10 \) does not work.

Thus, the only values of \( k \) that satisfy the condition are \( k = 1, 2, \) and \( 4 \).

The final answer is \( \boxed{1, 2, 4} \).