4. Find the smallest $n \in \mathbf{N}_{+}$, such that for any finite point set $M$ in the plane, if any $n$ points in $M$ can be covered by two lines, then there must exist two lines that can cover the point set $M$.
The required minimum value is 6.
When $|M| \leqslant 6$, the proposition is obviously true.
If $|M|>6$, for any six points $A_{1}, A_{2}, \cdots, A_{6}$ in $M$, they can be covered by two lines $l_{1}$ and $l_{2}$. Without loss of generality, assume that $l_{1}$ contains at least three points $A_{1}, A_{2}, A_{3}$. Let $M_{1}=M \cap l_{1}$ and $M_{2}=M \backslash l_{1}$.
If $\left|M_{2}\right| \leqslant 2$, the proposition is obviously true.
Without loss of generality, assume $\left|M_{2}\right| \geqslant 3$, and take $A_{1}, A_{2}, A_{3} \in M_{1}$, $B_{1}, B_{2}, B_{3} \in M_{2}$.

Then there exist two lines $l_{1}^{\prime}$ and $l_{2}^{\prime}$ that cover these six points. Among $A_{1}, A_{2}, A_{3}$, there must be two points on one of the lines, say $A_{1}, A_{2}$ are on $l_{1}^{\prime}$, then $l_{1}=l_{1}^{\prime}$. Points $B_{1}, B_{2}, B_{3}$ are not on $l_{1}^{\prime}$, so $B_{1}, B_{2}, B_{3}$ are on $l_{2}^{\prime}$. Replacing point $B_{3}$ with another point $B_{k}$ in $M_{2}$, then $B_{1}, B_{2}, B_{k}$ are collinear. Therefore, there exist two lines that can cover the point set $M$, and the proposition is true.

When $n=5$, there exists a counterexample. Take the three vertices and the midpoints of the three sides of a regular $\triangle ABC$. Any five of these points can be covered by two lines, but these six points cannot be covered by two lines.