Let $m$ and $n$ be positive integers. Find the smallest positive integer $s$ for which there exists an $m \times n$ rectangular array of positive integers such that
[list]
[*]each row contains $n$ distinct consecutive integers in some order,
[*]each column contains $m$ distinct consecutive integers in some order, and
[*]each entry is less than or equal to $s$.
[/list]

[i]Proposed by Ankan Bhattacharya.[/i]
1. **Initial Construction and Upper Bound:**
   We start by constructing an $m \times n$ array where each row contains $n$ distinct consecutive integers and each column contains $m$ distinct consecutive integers. We aim to show that the smallest possible $s$ for which such an array exists is $m + n - \gcd(m, n)$.

2. **Construction for $s = m + n - \gcd(m, n)$:**
   Let $\gcd(m, n) = d$. Consider a $d \times d$ block of consecutive integers. For a non-negative integer $k$, we can construct the block as follows:
   \[
   \begin{array}{cccc}
   k+1 & k+2 & \cdots & k+d \\
   k+2 & k+3 & \cdots & k+1 \\
   \vdots & \vdots & \ddots & \vdots \\
   k+d & k+1 & \cdots & k+d-1
   \end{array}
   \]
   We denote this block as $k^*$. Using these blocks, we can construct the $m \times n$ array as:
   \[
   \begin{array}{cccc}
   0^* & d^* & 2d^* & \cdots & (m-d)^* \\
   d^* & 2d^* & 3d^* & \cdots & m^* \\
   \vdots & \vdots & \vdots & \ddots & \vdots \\
   (n-d)^* & n^* & (n+d)^* & \cdots & (m+n-2d)^*
   \end{array}
   \]
   This construction ensures that each row and column contains distinct consecutive integers, and the largest number in the array is $m + n - d$.

3. **Proof of Minimality:**
   To show that $s = m + n - \gcd(m, n)$ is indeed the smallest possible value, we consider a working $m \times n$ grid for the triple $(m, n, s)$ and assume without loss of generality that $m \geq n$.

4. **Column Analysis:**
   Let $a_i$ denote the number of columns having the smallest number as $i$. Since each column contains $m$ distinct consecutive integers, the largest number in the grid is at most $m + n - 1$. For each row, the smallest number it contains should be at most $n$; otherwise, the grid would contain a number at least $n + 1 + (m - 1) = m + n$, which is a contradiction.

5. **Row Analysis:**
   For any positive integer $i$ with $n \leq i \leq m$, each row contains exactly one occurrence of $i$. Thus, there are exactly $n$ occurrences of $i$. The number of times $i$ occurs can be written as $a_i + a_{i-1} + \cdots + a_{i-n+1} = n$. This implies that the sequence $a_k$ repeats every $n$ terms.

6. **Reduction Process:**
   We remove the $n$ columns having the smallest number $\leq n$. For the remaining $(m-n) \times n$ grid, we subtract $n$ from each number. This process induces a working table for $(m-n, n, s-n)$ from the original table for $(m, n, s)$. By applying the Euclidean algorithm, we reduce the problem to a $d \times d$ grid where $d = \gcd(m, n)$.

7. **Final Step:**
   For a $d \times d$ grid, the smallest $s$ must be $d$ since each row has $d$ consecutive positive elements. Thus, $s + 2d - m - n \geq d$ or $s \geq m + n - d$. Since we already provided a construction for $s = m + n - \gcd(m, n)$, this is the smallest possible value.

\[
\boxed{m + n - \gcd(m, n)}
\]