Problem 1442

1. Understanding the Problem:
- We are given an integer $m > 1$ and an odd number $n$ such that $3 \le n < 2m$.
- We have a matrix $a_{i,j}$ with $1 \le i \le m$ and $1 \le j \le n$.
- The matrix satisfies two conditions:
1. For any $1 \le j \le n$, the elements $a_{1,j}, a_{2,j}, \ldots, a_{m,j}$ form a permutation of $1, 2, \ldots, m$.
2. For any $1 < i \le m$ and $1 \le j \le n-1$, $|a_{i,j} - a_{i,j+1}| \le 1$.
- We need to find the minimal value of $M$, where $M = \max_{1 < i < m} \sum_{j=1}^n a_{i,j}$.

2. Analyzing the Conditions:
- The first condition ensures that each column of the matrix is a permutation of $1, 2, \ldots, m$.
- The second condition ensures that the difference between consecutive elements in any row (except the first row) is at most 1.

3. Constructing the Matrix:
- To minimize $M$, we need to construct the matrix such that the sum of the elements in each row (except the first row) is as small as possible.
- Let's consider the simplest case where $m = 3$ and $n = 3$:
$$\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 3 & 3 & 1 \\ \end{array}$$
- Here, the sums of the rows are:
$$\sum_{j=1}^3 a_{1,j} = 1 + 2 + 3 = 6$$
$$\sum_{j=1}^3 a_{2,j} = 2 + 1 + 2 = 5$$
$$\sum_{j=1}^3 a_{3,j} = 3 + 3 + 1 = 7$$
- Therefore, $M = \max(5, 7) = 7$.

4. Generalizing the Construction:
- For a general $m$ and $n$, we can construct the matrix in a similar manner.
- For example, for $m = 4$ and $n = 3$:
$$\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 3 & 3 & 1 \\ 4 & 4 & 4 \\ \end{array}$$
- Here, the sums of the rows are:
$$\sum_{j=1}^3 a_{1,j} = 1 + 2 + 3 = 6$$
$$\sum_{j=1}^3 a_{2,j} = 2 + 1 + 2 = 5$$
$$\sum_{j=1}^3 a_{3,j} = 3 + 3 + 1 = 7$$
$$\sum_{j=1}^3 a_{4,j} = 4 + 4 + 4 = 12$$
- Therefore, $M = \max(5, 7, 12) = 12$.

5. **Finding the Minimal Value of $M$:**
- To minimize $M$, we need to ensure that the sums of the rows (except the first row) are as balanced as possible.
- The minimal value of $M$ can be achieved by constructing the matrix such that the sums of the rows are as close to each other as possible.

The final answer is $\boxed{7}$.