Find the smallest natural number $n$ that the following statement holds :
Let $A$ be a finite subset of $\mathbb R^{2}$. For each $n$ points in $A$ there are two lines including these $n$ points. All of the points lie on two lines.
1. **Identify the smallest number \( n \) such that any subset of \( n \) points in \( A \) can be covered by two lines:**

   We need to find the smallest natural number \( n \) such that for any finite subset \( A \subset \mathbb{R}^2 \), if any \( n \) points in \( A \) can be covered by two lines, then all points in \( A \) lie on two lines.

2. **Construct a counterexample for \( n = 5 \):**

   Consider the set \( E \) with the points \((0,0), (2,0), (2,2), (0,2), (0,1), (1,1)\). We need to verify that any subset of 5 points from \( E \) can be covered by two lines, but \( E \) itself cannot be covered by two lines.

   - Any subset of 5 points from \( E \) can be covered by two lines. For example:
     - \(\{(0,0), (2,0), (2,2), (0,2), (0,1)\}\) can be covered by the lines \( y = 0 \) and \( x = 0 \).
     - \(\{(0,0), (2,0), (2,2), (0,2), (1,1)\}\) can be covered by the lines \( y = x \) and \( y = 0 \).

   - However, the entire set \( E \) cannot be covered by two lines. Any covering of \( E \) by two lines would separate the six points into three collinear points each. The unique triple of collinear points from \( E \) containing \((0,1)\) is \((0,0), (0,1), (0,2)\). Therefore, the second line must cover the remaining three points \((2,0), (2,2), (1,1)\), which are not collinear.

3. **Prove that \( n = 6 \) has the desired property:**

   Let \( E \) be any set of at least 6 points in the plane such that any subset of 6 points from \( E \) can be covered by two lines. We need to show that all points in \( E \) lie on two lines.

   - Let \( F \) be any subset of \( E \) with 6 points. By the pigeonhole principle, at least three points from \( F \) must be collinear. Let these points be \( A, B, C \) on line \( \Delta \).

   - If \( E \) contains at most two points not on \( \Delta \), we are done. Otherwise, let \( X, Y \) be two points from \( E \) not on \( \Delta \), and let \( \Delta' \) be the line through \( X \) and \( Y \).

   - For any point \( M \) in \( E \) not equal to \( X \) or \( Y \) and not on \( \Delta \), the set \(\{A, B, C, X, Y, M\}\) can be covered by two lines. If neither of these lines is \( \Delta \), the line covering \( A \) cannot cover \( B \) or \( C \). Thus, the other line must cover \( B \) and \( C \), which means it is \( \Delta \), a contradiction.

   - Therefore, one of the lines must be \( \Delta \). It follows that the other line must cover \( X, Y, M \), forcing this second line to be \( \Delta' \) and \( M \in \Delta' \).

   - Thus, any point from \( E \) belongs either to \( \Delta \) or \( \Delta' \), proving that \( E \) can be covered by two lines.

\(\blacksquare\)

The final answer is \( \boxed{ n = 6 } \).