It is obvious that using $k$ lines, one can produce at most $\frac{k(k-1)}{2}$ points, so we claim that for all numbers $n$ such that $\frac{(k-1)k}{2} < n \leq \frac{k(k+1)}{2}$ the answer is $k+1$. We will prove it by induction on $k$. Consider all numbers from $\frac{(k-2)(k-1)}{2} + 1$ to $\frac{(k-1)k}{2}$ can be produced by $k$ lines; then we add a new line such that has an intersection with all previous lines in new points then we achieve the numbers $\frac{(k-1)k}{2} + 2$ to $\frac{k(k+1)}{2}$ using $k+1$ lines and it remains to show that $k+1$ lines can produce exactly $\frac{(k-1)k}{2} + 1$. For the remaining case, consider the sample that produces exactly $\frac{(k-2)(k-1)}{2} + 2$ intersection points and add a new line that has an intersection with $k-2$ lines in new points and passes through the intersection point of two other lines(by dual Sylvester theorem we can find these two points) , it solves the remaining case.