Let $1\le k\le n$ be integers. At most how many $k$-element subsets can we select from $\{1,2,\dots,n\}$ such that for any two selected subsets, one of the subsets consists of the $k$ smallest elements of their union?
1. **Initial Observation**:
   We start by noting that the set $\{ \{ 1,2,...,k-1,i\} \mid i=k,k+1,\dotsc ,n\}$ consists of $n-k+1$ $k$-element subsets that satisfy the given condition. We need to show that there cannot be more than $n-k+1$ such subsets.

2. **Assumption for Contradiction**:
   Suppose, for the sake of contradiction, that we can select $t > n-k+1$ such subsets. Let these $t$ subsets be $S_i = \{ a_{i,1}, a_{i,2}, \dotsc, a_{i,k} \}$ for $i = 1, 2, \dotsc, t$, where $a_{i,1} < a_{i,2} < \cdots < a_{i,k}$ and $\sum_{j=1}^{k} a_{p,j} \leq \sum_{j=1}^{k} a_{q,j}$ when $p < q \in \{ 1, 2, \dotsc, t \}$. This implies that $S_p$ contains the $k$ smallest elements of $S_p \cup S_q$ for all $p < q \in \{ 1, 2, \dotsc, t \}$.

3. **Distinctness of Elements**:
   For any $p < q \in \{ 1, 2, \dotsc, t \}$, we have $a_{p,k} \neq a_{q,k}$ (otherwise $\{ a_{p,1}, a_{p,2}, \dotsc, a_{p,k} \} = \{ a_{q,1}, a_{q,2}, \dotsc, a_{q,k} \}$). Also, we have $a_{p,k} \leq a_{q,k}$ (otherwise $a_{p,k}$ is not in the $k$ smallest elements of $S_p \cup S_q$). Therefore, $a_{1,k} < a_{2,k} < \cdots < a_{t,k}$.

4. **Bounding the Largest Element**:
   From the above, we get $a_{t,k} \geq t-1 + a_{1,k}$. Combining this with the fact that $a_{1,1} < a_{1,2} < \cdots < a_{1,k}$, we get $a_{1,k} \geq a_{1,1} + k - 1 \geq 1 + k - 1 = k$. Thus, $a_{t,k} \geq t-1 + k > (n-k+1) - 1 + k = n$.

5. **Contradiction**:
   The inequality $a_{t,k} \geq n$ contradicts the fact that $a_{t,k}$ should be an element of the set $\{1, 2, \dotsc, n\}$. Hence, our assumption that $t > n-k+1$ is false.

Therefore, the maximum number of $k$-element subsets that can be selected from $\{1, 2, \dotsc, n\}$ such that for any two selected subsets, one of the subsets consists of the $k$ smallest elements of their union, is $n-k+1$.

The final answer is $\boxed{n-k+1}$