Problem 1805

Given a positive integer $n \ge 2$, we aim to find all $n$-tuples of positive integers $(a_1, a_2, \ldots, a_n)$ such that $1 < a_1 \le a_2 \le a_3 \le \cdots \le a_n$, $a_1$ is odd, and the following conditions hold:
1. $M = \frac{1}{2^n}(a_1-1)a_2 a_3 \cdots a_n$ is a positive integer.
2. One can pick $n$-tuples of integers $(k_{i,1}, k_{i,2}, \ldots, k_{i,n})$ for $i = 1, 2, \ldots, M$ such that for any $1 \le i_1 < i_2 \le M$, there exists $j \in \{1, 2, \ldots, n\}$ such that $k_{i_1, j} - k_{i_2, j} \not\equiv 0, \pm 1 \pmod{a_j}$.

To solve this, we first note the necessary condition for $M$ to be a positive integer:
$$2^n \mid (a_1 - 1).$$

This means $a_1 - 1$ must be divisible by $2^n$. Given $a_1$ is odd, we can write $a_1$ as:
$$a_1 = k \cdot 2^n + 1$$
for some positive integer $k$.

Next, we need to ensure that the $n$-tuples $(k_{i,1}, k_{i,2}, \ldots, k_{i,n})$ can be chosen such that the second condition is satisfied. This can be achieved by constructing the points in a specific manner.

For simplicity, consider the case where all $a_i$ are equal and odd:
$$a_1 = a_2 = \cdots = a_n = k \cdot 2^n + 1.$$

We can place points on the line $x_2 = c_2, x_3 = c_3, \ldots, x_n = c_n$ for fixed $c_2, \ldots, c_n$ such that:
$$x_1 \in \{0, 2, \ldots, 2(k-1)\} + 2k \sum_{j=1}^{n-1} 2^{j-1} c_{j+1}.$$

This ensures that the points are distributed in a way that satisfies the second condition. By extending this construction to cases where $a_2, \ldots, a_n$ are odd integers at least $a_1$, we can generalize the solution.

Thus, the $n$-tuples $(a_1, a_2, \ldots, a_n)$ that satisfy the given conditions are those where $a_1 = k \cdot 2^n + 1$ for some positive integer $k$, and $a_2, \ldots, a_n$ are odd integers such that $1 < a_1 \le a_2 \le \cdots \le a_n$.

The answer is: $\boxed{(a_1, a_2, \ldots, a_n) \text{ where } a_1 = k \cdot 2^n + 1 \text{ and } a_2, \ldots, a_n \text{ are odd integers such that } 1 < a_1 \le a_2 \le \cdots \le a_n}$.