Three. (20 points) A tetrahedron has 6 edges and 4 triangular faces. Given that the 6 edge lengths are exactly the positive integers $n, n+$ $1, n+2, n+3, n+4, n+5$. If the perimeter of a certain triangular face is a multiple of 3, then this triangle is painted red; otherwise, it is painted yellow. Question: What is the maximum number of yellow triangles? Prove your conclusion.
Three, at most 3 yellow triangles.
Since 6 consecutive positive integers divided by 3 must have remainders of 2 zeros, 2 ones, and 2 twos, denoted as $a, a, b, b, c, c$. If there are 4 yellow triangles, take any yellow $\triangle ABC$, then it must have two sides with the same remainder when divided by 3, denoted as $a, a$ for sides $AB \equiv AC \equiv a(\bmod 3)$. Since the two $a$s are already used, $AD$ must not have a remainder of $a$ when divided by 3, denoted as $AD \equiv b(\bmod 3)$, then among $BD, CD$,
at most one can have a remainder of $b$ when divided by 3 (as shown in Figure 10).
When $BD \equiv c(\bmod 3)$

$\triangle ABD$ is a red triangle;
When $CD \equiv c(\bmod 3)$

$\triangle ACD$ is a red triangle.
Therefore, there cannot be 4 yellow triangles.
Furthermore, when $AB=3, BC=4, CA=5, AD=6, CD=8, BD=7$, except for $\triangle ABC$, there are 3 yellow triangles.
Thus, the maximum number of yellow triangles is 3.
(Luo Zengru, Department of Mathematics, Shaanxi Normal University, 710062)