In a pile you have 100 stones. A partition of the pile in $ k$ piles is [i]good[/i] if:

1) the small piles have different numbers of stones;

2) for any partition of one of the small piles in 2 smaller piles, among the $ k \plus{} 1$ piles you get 2 with the same number of stones (any pile has at least 1 stone).

Find the maximum and minimal values of $ k$ for which this is possible.
1. **Determine the upper bound for \( k \):**
   - If \( k \geq 14 \), then the sum of the first \( k \) natural numbers is:
     \[
     1 + 2 + \ldots + 14 = \frac{14 \cdot 15}{2} = 105 > 100
     \]
     This is a contradiction since we only have 100 stones. Therefore, \( k \leq 13 \).

2. **Verify if \( k = 13 \) is possible:**
   - Consider the set \( \{1, 2, \ldots, 11, 12, 22\} \). The sum of these numbers is:
     \[
     1 + 2 + 3 + \ldots + 11 + 12 + 22 = \frac{11 \cdot 12}{2} + 12 + 22 = 66 + 12 + 22 = 100
     \]
     This satisfies the condition of having 100 stones. Additionally, any partition of one of these piles into two smaller piles will result in at least two piles having the same number of stones. Therefore, \( k = 13 \) is possible.

3. **Verify if \( k = 10 \) is possible:**
   - Consider the set \( \{1, 3, 5, \ldots, 19\} \). The sum of these numbers is:
     \[
     1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
     \]
     This satisfies the condition of having 100 stones. Additionally, any partition of one of these piles into two smaller piles will result in one odd and one even number, ensuring that the odd number is among the smaller piles. Therefore, \( k = 10 \) is possible.

4. **Show that \( k < 10 \) does not work:**
   - Suppose there is no 1-stone pile. If there is a pile with \( n \) stones, there must be one with \( n-1 \) stones because \( n = (n-1) + 1 \). Therefore, if \( n \) is the number of stones in the biggest pile, the total number of stones is:
     \[
     n + (n-1) + \ldots + 2 = \frac{n(n+1)}{2} - 1
     \]
     This cannot be 100. Hence, there must be a pile with just 1 stone.

5. **Analyze the case when the biggest pile has \( 2m+1 \) stones:**
   - At least one of the pairs \( (1, 2m), (2, 2m-1), \ldots, (m, m+1) \) must be present. Therefore, there are at least \( m+1 \) piles in total.

6. **Analyze the case when the biggest pile has \( 2m \) stones:**
   - There are at least \( m \) piles.

7. **Show that \( k \leq 8 \) does not work:**
   - If \( k \leq 8 \), the biggest pile has at most \( 2k = 16 \) stones. The total number of stones is:
     \[
     1 + 16 + 15 + \ldots + 10 = 92 < 100
     \]
     This is a contradiction. Therefore, \( k \geq 9 \).

8. **Analyze the case when \( k = 9 \):**
   - The biggest pile has at most 18 stones. Consider the smallest pile greater than 10. By the previous analysis, there are at least 5 piles less than or equal to 10. If the biggest pile has 17 or fewer stones, the total number of stones is:
     \[
     1 + 7 + 8 + 9 + 10 + 14 + 15 + 16 + 17 = 97 < 100
     \]
     This is a contradiction. Therefore, the biggest pile must have 18 stones. However, the smallest pile greater than 1 can be at most 4. The total number of stones is:
     \[
     1 + 4 + 8 + 9 + 10 + 15 + 16 + 17 + 18 = 98 < 100
     \]
     This is a contradiction. Therefore, \( k > 9 \).

Conclusion:
The maximum value of \( k \) is 13, and the minimum value of \( k \) is 10.

The final answer is \( \boxed{ k_{\text{max}} = 13 } \) and \( k_{\text{min}} = 10 \).