13. Let $A$ be a set of 6 points consisting of the three vertices and the midpoints of the three sides of a triangle. Then any 5 points in $A$ can be covered by 2 lines, but the 6 points in $A$ cannot be covered by any two lines. Therefore, the smallest positive integer $n$ $\geqslant 6$. Next, assume that any 6 points in $A$ can be covered by 2 lines. Take any 6 points in $A$, then there must be 2 lines $l_{1}$ and $l_{2}$ that cover these 6 points, so one of $l_{1}$ and $l_{2}$ (denoted as $l_{1}$) must cover at least three points $A_{1}, A_{2}, A_{3}$ in $A$. Let the set of points in $A$ except those on $l_{1}$ be denoted as $S$. If $|S| \leqslant 2$, then all points in $S$ can be covered by another line $l^{\prime}$. If $|S| \geqslant 3$, then take any 3 points $B_{1}, B_{2}, B_{3}$ in $S$, then $A_{1}, A_{2}, A_{3}, B_{1}, B_{2}, B_{3}$ can be covered by 2 lines, so one of them must cover at least 2 points among $A_{1}, A_{2}, A_{3}$, hence this line can only be $l_{1}$. Let the other line be $l^{\prime}$, then $B_{1}, B_{2}, B_{3}$ are all on $l^{\prime}$, i.e., any 3 points in $S$ are on the same line $l^{\prime}$, so all points in $S$ are on the same line $l^{\prime}$, i.e., the two lines $l_{1}$ and $l^{\prime}$ cover all points in $A$. In summary, the smallest positive integer $n=6$.
13. Find the smallest positive integer $n$ such that: if for any finite set of points $A$ in the plane, any $n$ points can be covered by two lines, then there exist two lines that can cover all points in $A$.