Three. (This question is worth 50 points) Let $S=\{1,2,3, \cdots, 100\}$. Find the largest integer $k$, such that $S$ has $k$ distinct non-empty subsets, with the property that for any two different subsets among these $k$ subsets, if their intersection is non-empty, then the smallest element in their intersection is different from the largest elements in both of these subsets.

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For a finite non-empty set of real numbers $A$, use $\min A$ and $\max A$ to denote the smallest and largest elements of $A$, respectively. Consider all subsets of $S$ that contain 1 and have at least two elements, a total of $2^{99}-1$ such subsets. They clearly meet the requirement, because $\min \left(A_{i} \cap A_{j}\right)=1<\max A_{i}$. Hence, $k_{\max } \geq 2^{99}-1$. 10 points

Next, we prove that when $k \geq 2^{99}$, there do not exist $k$ subsets that satisfy the requirement. We use mathematical induction to prove: for an integer $n \geq 3$, in any $m\left(\geq 2^{n-1}\right)$ different non-empty subsets $A_{1}, A_{2}, \cdots, A_{m}$ of the set $\{1,2, \cdots, n\}$, there exist two subsets $A_{i}, A_{j}, i \neq j$, such that
$$
A_{i} \cap A_{j} \neq \varnothing \text {, and } \min \left(A_{i} \cap A_{j}\right)=\max A_{i} \text {. }
$$

It is clear that we only need to prove the above conclusion for $m=2^{n-1}$.
When $n=3$, divide all 7 non-empty subsets of $\{1,2,3\}$ into 3 groups: the first group: $\{3\},\{1,3\}$, $\{2,3\}$; the second group: $\{2\},\{1,2\}$; the third group: $\{1\},\{1,2,3\}$. By the pigeonhole principle, any 4 non-empty subsets must have two in the same group. Taking the two subsets in the same group and denoting them as $A_{i}, A_{j}$, with the one listed first as $A_{2}$, they satisfy (1). 20 points
Assume the conclusion holds for $n (\geq 3)$, and consider the case for $n+1$. If at least $2^{n-1}$ of the subsets $A_{1}, A_{2}, \cdots, A_{2^{n}}$ do not contain $n+1$, apply the induction hypothesis to these $2^{n-1}$ subsets, and we know there exist two subsets that satisfy (1). $\qquad$
If at most $2^{n-1}-1$ subsets do not contain $n+1$, then at least $2^{n-1}+1$ subsets contain $n+1$. Removing $n+1$ from these $2^{n-1}+1$ subsets, we get $2^{n-1}+1$ subsets of $\{1,2, \cdots, n\}$.

Since all subsets of $\{1,2, \cdots, n\}$ can be divided into $2^{n-1}$ groups, each group consisting of two complementary subsets, by the pigeonhole principle, among the above $2^{n-1}+1$ subsets, there must be two in the same group, i.e., they are complements of each other. Therefore, there are two subsets $A_{i}, A_{j}$ such that $A_{i} \cap A_{j}=\{n+1\}$, which clearly satisfy (1). Hence, the conclusion holds for $n+1$.
In summary, the required $k_{\max }=2^{99}-1$.
.50 points