It's easy to see that the number of regions formed by the red lines is $n(2n+1)+1$, each blue line cuts exactly $2n+1$ of those regions, then after putting $n$ blue lines, there is at least one region with red border. But, if there were only one, all the blue lines cut different regions, and in particular they don't intersect, that's impossible, so there are at least two. In fact, there are at least $n$ regions with red border, because the first blue line cuts $2n+1$ regions, but the second must intersect the other blue line in some region, so it cuts at most $2n$ regions with red border, and the same happens for the other $n-2$ blue lines. So, in total they cut at most $n(2n+1)+1-n$ regions with red border. The result is still true if we suppose that each blue line is in general position respect to the red lines, but not necessarily respect to the others blue lines. In fact, if two blue lines are parallel, they cut the same two unbounded regions with red border.

Never mind I'm dumb

Zaratustra wrote: In fact, there are at least $n$ regions with red border, because the first blue line cuts $2n+1$ regions, but the second must intersect the other blue line in some region, so it cuts at most $2n$ regions with red border, and the same happens for the other $n-2$ blue lines. So, in total they cut at most $n(2n+1)+1-n$ regions with red border. The result is still true if we suppose that each blue line is in general position respect to the red lines, but not necessarily respect to the others blue lines. In fact, if two blue lines are parallel, they cut the same two unbounded regions with red border. Can anyone assert whether the above is true? I seem to have some doubts about it, but I'm waiting for the assertion first.

It's easy to show (induction!) that $3n$ lines generate ${3n+1 \choose 2} + 1$ different regions. Let's say that a region is good if all of its borders are red, and lets try to count the number of bad regions. Let's focus on a blue line, which is cut into exactly $3n$ (induction!) segments. Each one of these segments touch exactly two regions, making them bad regions. Since there are $n$ blue lines, we have $n \cdot 3n \cdot 2 = 6n^2$ bad regions. However, whenever two blue lines meet, they count neighboring regions twice at their intersection. We should remove such regions. There are ${n\choose2}$ such blue-blue intersections, and they are associated to $4$ regions, which gives $2n(n-1) = 2n^2-2n$. So, the number of bad regions is at most $$6n^2-(2n^2-2n) = 4n^2+2n$$ The minimum for the number of good regions is therefore $${3n+1 \choose 2} + 1 - (4n^2 + 2n)= \frac{9n^2}{2} + \frac{3n}{2} + 1 - (4n^2 + 2n) = \frac{n^2}{2} - \frac{n}{2} + 1$$ Which is at least $2$ for $n \geq 2$