Three. (50 points) Let $S=\{1,2, \cdots, 100\}$. Find the largest integer $k$, such that the set $S$ has $k$ distinct non-empty subsets with the property: 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 set $A$, respectively.

Consider all subsets of set $S$ that contain 1 and have at least two elements.
Notice that, $\min \left(A_{i} \cap A_{j}\right)=1<\max A_{i}$.
Therefore, $k_{\max } \geqslant 2^{99}-1$.
Thus, there are $2^{99}-1$ such subsets.
It is clearly satisfactory.
Next, we prove that when $k \geqslant 2^{99}$, there do not exist $k$ subsets that meet the requirements.

We use mathematical induction to prove: For integers $n \geqslant 3$, in any $m\left(m \geqslant 2^{n-1}\right)$ 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 $A_{i} \cap A_{j} \neq \varnothing$, and $\min \left(A_{i} \cap A_{j}\right)=\max A_{i}$.
Clearly, it is only necessary to prove the above conclusion for $m=2^{n-1}$.
When $n=3$, divide all seven non-empty subsets of $\{1,2,3\}$ into three groups:
First group: $\{3\},\{1,3\},\{2,3\}$;
Second group: $\{2\},\{1,2\}$;
Third group: $\{1\},\{1,2,3\}$.
By the pigeonhole principle, any four non-empty subsets must have two in the same group. Taking the two subsets in the same group and denoting the one that comes first as $A_{i}$, the conclusion (1) is satisfied.

Assume the conclusion holds for $n(n \geqslant 3)$. 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 conclusion (1).

If at most $2^{n-1}-1$ of the subsets do not contain $n+1$, then at least $2^{n-1}+1$ of the subsets contain $n+1$. Removing $n+1$ from these $2^{n-1}+1$ subsets, we obtain $2^{n-1}+1$ subsets of $\{1,2, \cdots, n\}$.

Since all subsets of $\{1,2, \cdots, n\}$ can be divided into $2^{n-1}$ groups, each containing 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 $A_{i} \cap A_{j}=\{n+1\}$, and these two subsets clearly satisfy conclusion (1).
Thus, the conclusion holds for $n+1$.
In summary, $k_{\max }=2^{99}-1$.