Let the set of 12 red points be $A = \{A_1, A_2, \dots, A_{12}\}$. From $A_1$, one can get 11 chords with red points, but every triangle with vertex $A_1$ has two such chords, and so the 11 chords should be in at least 6 triangles with $A_1$ being an endpoint. The same applies to $A_i$ ($i = 2, 3, \dots, 12$); we need

$12 \times 6 = 72$ triangles, and every triangle has 3 points. So $n \ge \frac{72}{3} = 24$.

On the other hand, the following example shows that $n$ can be 24.

Consider a circle whose perimeter is 12, and the 12 red points are on the circle with equal distance.

The number of chords with red endpoint is $\binom{12}{2} = 66$. If the length of the minor arc to a chord is $k$, then say “the chord belongs to $k$.” So we have only six kinds of chords, and the number of chords belonging to 1, 2, ..., 5 is 12, and that belonging to 6 is 6.

If three chords belonging to $a, b, c$ ($a \le b \le c$) can be the sides of a triangle, then $a + b = c$ or $a + b + c = 12$. So the triangles $(a, b, c) \in \{(1, 1, 2), (2, 2, 4), (3, 3, 6), (2, 5, 5), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6), (3, 4, 5), (4, 4, 4)\}$.

Now we give an example:

The number of (1, 2, 3) triangles is 6, whose vertices are
$\{2, 3, 5\}, \{4, 5, 7\}, \{6, 7, 9\}, \{8, 9, 11\}, \{10, 11, 1\}, \{12, 1, 3\}$.

The number of (1, 5, 6) triangles is 6, whose vertices are
$\{1, 2, 7\}, \{3, 4, 9\}, \{5, 6, 11\}, \{7, 8, 1\}, \{9, 10, 3\}, \{11, 12, 5\}$.

The number of (2, 3, 5) triangles is 6, whose vertices are
$\{2, 4, 11\}, \{4, 6, 1\}, \{6, 8, 3\}, \{8, 10, 5\}, \{10, 12, 7\}, \{12, 2, 9\}$.

The number of (4, 4, 4) triangles is 3, whose vertices are
$\{1, 5, 9\}, \{2, 6, 10\}, \{3, 7, 11\}$.

The number of (2, 4, 6) triangles is 3, whose vertices are
$\{4, 6, 12\}, \{8, 10, 4\}, \{12, 2, 8\}$.

So the minimum $n$ is 24.