3. 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 subsets.
First, consider the subsets of $S$ that are different from $\{1\}$ and contain the element 1, there are $2^{99}-1$ such subsets. For any of these subsets, they must contain at least one number from $2,3, \cdots, 100$, and their maximum element is certainly not 1. Therefore, for any two of these subsets, the smallest element of their intersection is 1, which is different from the maximum elements of these two subsets, hence $k \geqslant 2^{99}-1$.

Next, let $S_{k}$ be the set of these $k$ subsets.
If $\{1\} \in S_{k}$, then all subsets containing 1 cannot be in $S_{k}$, otherwise the smallest element of the intersection of any two such subsets would be 1, and the maximum element of $\{1\}$ is also 1, which is a contradiction. Therefore, the remaining elements in $S_{k}$ can only be non-empty subsets of the set $\{2,3, \cdots, 100\}$. Clearly, $\{2\}$ and $\{2,3\}$ cannot both be in $S_{k}$, so $k-1 \leqslant 2^{99}-2 \Rightarrow k \leqslant 2^{99}-1$;
If $\{1\} \notin S_{k}$, consider the subsets in $S_{k}$ whose maximum element is 100. For any such subset $\left\{100, a_{1}, a_{2}, \cdots, a_{n}\right\}$, then $\{100\} \bigcup\left(\{1,2, \cdots, 99\} \backslash\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\right) \notin S_{k}$, otherwise the intersection of these two subsets would be $\{100\}$, and their respective maximum elements would also be 100, which is a contradiction. Therefore, the number of subsets in $S_{k}$ whose maximum element is 100 is at most half the number of subsets of the set $\{1,2,3, \cdots, 99\}$, i.e., $2^{98}$; similarly, the number of subsets in $S_{k}$ whose maximum element is $i$ is at most half the number of subsets of the set $\{1,2,3, \cdots, i-1\}$, i.e., $2^{i-2}, i=2,3, \cdots, 100$.
Since $\{1\} \notin S_{k}$, we have $k=\left|S_{k}\right| \leqslant \sum_{i=2}^{100} 2^{i-2}=2^{99}-1$.

In summary, the largest integer $k$ that satisfies the problem is $2^{99}-1$.