Let $S=\{1,2,3,\cdots,100\}$. Find the maximum value of integer $k$, such that there exist $k$ different nonempty subsets of $S$ satisfying the condition: for any two of the $k$ subsets, if their intersection is nonemply, then the minimal element of their intersection is not equal to the maximal element of either of the two subsets.
1. **Define the set and the problem:**
   Let \( S = \{1, 2, 3, \ldots, 100\} \). We need to find the maximum value of integer \( k \) such that there exist \( k \) different nonempty subsets of \( S \) satisfying the condition: for any two of the \( k \) subsets, if their intersection is nonempty, then the minimal element of their intersection is not equal to the maximal element of either of the two subsets.

2. **Construct a candidate solution:**
   Consider all subsets of \( S \) that contain the element \( 1 \), except for the subset \(\{1\}\). There are \( 2^{99} - 1 \) such subsets because each subset of \( S \) containing \( 1 \) can be formed by choosing any subset of \( \{2, 3, \ldots, 100\} \), and there are \( 2^{99} \) such subsets in total, including the empty subset. Excluding the subset \(\{1\}\), we have \( 2^{99} - 1 \) subsets.

3. **Verify the condition:**
   For any two subsets \( A \) and \( B \) in our collection, both contain \( 1 \). If their intersection is nonempty, it must contain \( 1 \). The minimal element of their intersection is \( 1 \), which is not the maximal element of either \( A \) or \( B \) (since \( A \) and \( B \) are not \(\{1\}\)). Thus, the condition is satisfied.

4. **Prove optimality using induction:**
   We need to show that \( 2^{99} - 1 \) is the maximum number of such subsets. We will use induction on \( n \), the cardinality of \( S \).

   - **Base case \( n = 2 \):**
     For \( S = \{1, 2\} \), the subsets are \(\{1\}, \{2\}, \{1, 2\}\). The pairs \((\{2\}, \{1, 2\})\) form a "horrible" pair because their intersection \(\{2\}\) is the maximal element of \(\{2\}\). Thus, we can only choose one subset from each pair, giving us \( 2^{1} - 1 = 1 \) subset.

   - **Inductive step:**
     Assume the statement holds for \( n = k \). For \( n = k+1 \), consider the set \( S = \{1, 2, \ldots, k+1\} \). By the induction hypothesis, we can pair all nonempty subsets of \( \{1, 2, \ldots, k\} \) into "horrible" pairs. We need to pair all nonempty subsets of \( S \) containing \( k+1 \) into "horrible" pairs. For any subsets \( A, B \) of \( \{1, 2, \ldots, k\} \) such that \( A \cup B = \{1, 2, \ldots, k\} \) and \( |A| + |B| = k \), we can pair \( \{k+1\} \cup A \) with \( \{k+1\} \cup B \). This completes the induction.

5. **Conclusion:**
   Since we can pair all nonempty subsets of \( S \) except for \(\{1\}\) into "horrible" pairs, we can pick at most one subset from each pair. This gives us an upper bound of \( 1 + \frac{2^{100} - 2}{2} = 2^{99} \). However, to achieve equality, we need \(\{1\}\) to be chosen, which would prevent any other subset containing \( 1 \) from being chosen. Thus, the maximum number of subsets is \( 2^{99} - 1 \).

The final answer is \( \boxed{ 2^{99} - 1 } \).