The answer is $4044$.

After Alice has drawn the $4$ red vertices of a square, we can draw one of the diagonals of the square to form $2$ triangles. Each time when Alice draws one more red point inside the square, it must lie inside one of the triangles formed. By joining the new red point to the $3$ vertices of this triangle, we obtain a new triangulation with $2$ more triangles. At the end, when Alice has drawn all the red points, there is a triangulation with $2 + 2 \times 2021 = 4044$ triangles. Since there is a blue point in each of these triangles, we must have $n \ge 4044$.

Next, we provide a strategy for Bob so that it is sufficient to draw $4044$ blue points. We set up a coordinate system such that all red points have different $x$-coordinates. For each red point $P(x, y)$ inside the square, Bob draws two blue points $(x, y+\epsilon)$ and $(x, y-\epsilon)$ where $2\epsilon$ is the distance from $P$ to the closest line joining two red points different from $P$. We say that these two blue points are associated to the point $P$. For the $4$ red vertices of the square, Bob does the same thing except that points outside the square need not be drawn. Note that there are only $2$ blue points each of which is associated to a vertex of the square whose $x$-coordinate is neither the smallest nor the largest. In total, there are $2 \times 2021 + 2 = 4044$ blue points. We claim that the condition is satisfied.

Indeed, consider any triangle $ABC$ with red vertices where the $x$-coordinates of $A$, $B$ and $C$ are in ascending order. WLOG assume $B$ lies below the line $AC$. Then $\angle ABC$ contains the blue point associated to $B$ lying above $B$ since this blue point also lies below the line $AC$. This completes the proof.