4. Given twelve red points on a circle. Find the minimum value of $n$, such that there exist $n$ triangles with red points as vertices, so that every chord with red points as endpoints is a side of one of these triangles.
（Supplied by Tao Pingsheng）
4. Let the set of red points be $A=\left\{A_{1}, A_{2}, \cdots, A_{12}\right\}$.

There are 11 chords passing through point $A_{1}$, and any triangle containing vertex $A_{1}$ contains exactly two chords passing through point $A_{1}$. Therefore, these 11 chords passing through point $A_{1}$ must be distributed among at least six triangles containing vertex $A_{1}$.

Similarly, the chords passing through point $A_{i} (i=2,3, \cdots, 12)$ must also be distributed among six triangles containing vertex $A_{i}$. This requires $12 \times 6=72$ triangles, and since each triangle has three vertices, they are counted three times. Therefore, at least $\frac{72}{3}=24$ triangles are needed.
Next, we prove that the lower bound of 24 can be achieved.
Without loss of generality, consider a circle with a circumference of 12, and its 12 equally divided points are red points. The number of chords with red endpoints is $\mathrm{C}_{12}^{2}=66$. If a chord subtends an arc of length $k$, we call the chord's scale $k$. Thus, there are only six scales for chords with red endpoints, where chords with scales of $1,2, \cdots, 5$ each have 12 chords, and chords with a scale of 6 have 6 chords.

This results in 24 triangles, and they satisfy the conditions of the problem.
Therefore, the minimum value of $n$ is 24.
If the scales of the chords forming the three sides of a triangle are $a, b, c (a \leqslant b \leqslant c)$, then either $a+b=c$ or $a+b+c=12$ must be satisfied.

Thus, there are only 12 possible combinations of scales for the sides of red point triangles:
$$
\begin{array}{l}
(a, b, c) \\
=(1,1,2),(2,2,4),(3,3,6),(2,5,5), \\
\quad(1,2,3),(1,3,4),(1,4,5),(1,5,6), \\
\quad(2,3,5),(2,4,6),(3,4,5),(4,4,4) .
\end{array}
$$

Here is one possible combination:
Take six each of the $(1,2,3)$, $(1,5,6)$, and $(2,3,5)$ types, and four of the $(4,4,4)$ type. In this case, exactly 66 chords are obtained, and among them, there are 12 chords each with scales of $1,2, \cdots, 5$, and 6 chords with a scale of 6.

Now, construct as follows: First, make six triangles each of the $(1,2,3)$, $(1,5,6)$, and $(2,3,5)$ types, and three triangles of the $(4,4,4)$ type, then supplement with three triangles of the $(2,4,6)$ type.
$(1,2,3)$ type six: Their vertex labels are
$\{2,3,5\},\{4,5,7\},\{6,7,9\}$,
$\{8,9,11\},\{10,11,1\},\{12,1,3\}$;
$(1,5,6)$ type six: Their vertex labels are
$\{1,2,7\},\{3,4,9\},\{5,6,11\}$,
$\{7,8,1\},\{9,10,3\},\{11,12,5\}$;
$(2,3,5)$ type six: Their vertex labels are
$\{2,4,11\},\{4,6,1\},\{6,8,3\}$,
$\{8,10,5\},\{10,12,7\},\{12,2,9\}$;
$(4,4,4)$ type three: Their vertex labels are
$\{1,5,9\},\{2,6,10\},\{3,7,11\}$;
$(2,4,6)$ type three: Their vertex labels are
$\{4,6,12\},\{8,10,4\},\{12,2,8\}$.
(The remaining triangles in each case can be obtained by rotating one of the triangles around the center of the circle as appropriate.)