To find the minimum positive integer $n \geq 3$ such that there exist $n$ points $A_1, A_2, \ldots, A_n$ satisfying no three points are collinear and for any $1 \leq i \leq n$, there exists $1 \leq j \leq n$ (with $j \neq i$), such that the segment $A_jA_{j+1}$ passes through the midpoint of segment $A_iA_{i+1}$, where $A_{n+1} = A_1$, we proceed as follows:

First, it is necessary to verify that $n = 3$ and $n = 4$ do not satisfy the given conditions. Through geometric construction and analysis, it can be shown that no such configurations exist for these values of $n$.

Next, consider $n = 5$. We analyze two cases:
1. Case 1: There are no parallelograms formed by any four of the points $A_i$. By detailed geometric analysis and coordinate bashing, it can be shown that no such five points exist.
2. Case 2: Assume $A_1A_4A_2A_3$ forms a parallelogram. By considering the reflection of points and ensuring no three points are collinear, it leads to a contradiction, proving that $n = 5$ is also not possible.

Finally, for $n = 6$, a construction exists that satisfies all the given conditions. Therefore, the minimum positive integer $n$ for which the conditions hold is $n = 6$.

The answer is: \boxed{6}.