Let $1 \leq k \leq n$ be integers. How many $k$-element subsets of the set $\{1,2, \ldots, n\}$ can be chosen such that any two of them consist of the $k$ smallest elements of their union?
Let $\mathcal{H}$ be a system of subsets with the properties described in the problem, and let $A$ and $B$ be distinct members of $\mathcal{H}$. Suppose that the $k$ smallest elements of the set $A \cup B$ form the set $A$. Given that $|A \cup B| > k$, the largest element of $A \cup B$ does not belong to $A$. Therefore, this element belongs to $B$, which means that the largest elements of the members of the family $\mathcal{H}$ are pairwise distinct. Thus, the number of sets in the family $\mathcal{H}$ can be at most as large as the number of different possible largest elements of a $k$-element subset of $\{1, 2, \ldots, n\}$. It is clear that none of the numbers $1, 2, \ldots, k-1$ can be the largest element of such a $k$-element subset, so the number of subsets in question can be at most the size of the set $\{k, k+1, \ldots, n\}$, which is $n - (k-1) = n - k + 1$.

To show that this upper bound is achievable, it suffices to notice that the sets $A_{i} := \{j \in \mathbb{N} : i \leq j \leq i + k - 1\}$ have the property described in the problem if $i \in \{1, 2, \ldots, n - k + 1\}$. This gives us exactly $n - k + 1$ sets, so the answer to the problem is $n - k + 1$.

Remark. Another example of a maximal-sized family, different from the one above, is the system of sets $\{1, 2, \ldots, k-1, i\}$ for $i \in \{k, k+1, \ldots, n\}$.