Finitely many polygons are placed in the plane. If for any two polygons of them, there exists a line through origin $O$ that cuts them both, then these polygons are called "properly placed". Find the least $m \in \mathbb{N}$, such that for any group of properly placed polygons, $m$ lines can drawn through $O$ and every polygon is cut by at least one of these $m$ lines.
1. **Initial Observation**: We need to find the smallest number \( m \) such that for any group of properly placed polygons, \( m \) lines can be drawn through the origin \( O \) and every polygon is cut by at least one of these \( m \) lines.

2. **Case \( m = 1 \)**: 
   - Assume \( m = 1 \). 
   - Consider a line \( AB \) through \( O \) with points \( A \) and \( B \) on opposite sides of \( O \).
   - Now, consider rays \( OC \) and \( OD \) on the same side of \( AB \), with \( OC \) between \( OB \) and \( OD \).
   - We can construct polygons \( BC \), \( CD \), and \( DA \) such that each polygon is on a different side of the line \( AB \).
   - This configuration shows that a single line cannot cut all polygons, hence \( m = 1 \) does not work.

3. **Case \( m = 2 \)**:
   - For each polygon in the collection, consider the minimal angular region between two lines through \( O \) that contains the polygon.
   - Select a polygon \( \mathcal{P} \) whose region is minimal with respect to inclusion, meaning no other polygon's region is properly included within \( \mathcal{P} \)'s region.
   - The two lines through \( O \) that bound the region of \( \mathcal{P} \) will cut \( \mathcal{P} \).
   - Since the polygons are properly placed, any other polygon must intersect at least one of these two lines, as otherwise, it would contradict the minimality of \( \mathcal{P} \)'s region.
   - Therefore, \( m = 2 \) lines are sufficient to ensure that every polygon is cut by at least one of these lines.

4. **Conclusion**: We have shown that \( m = 1 \) does not work and \( m = 2 \) works. Therefore, the least \( m \) such that for any group of properly placed polygons, \( m \) lines can be drawn through \( O \) and every polygon is cut by at least one of these \( m \) lines is \( m = 2 \).

The final answer is \( \boxed{ m = 2 } \)