Let \(k\) be the number of elements of the first color. Then the number of elements of the other color is \(n-k\). Then the maximum number of non-monochromatic three-element subsets of \(M_n\) is obviously \[\binom{k}{2}(n-k) + \binom{n-k}{2}k = \frac{k(n-k)(n-2)}{2}.\]The last expression attains its maximum for \(\displaystyle k=\left\lfloor \frac{n}{2}\right\rfloor\). Therefore the minimal number of three-element subsets of \(M_n\) for which for any coloring there is a monochrmonatic subset is \(\displaystyle\frac{1}{2} \left\lfloor \frac{n}{2}\right\rfloor \left(n - \left\lfloor \frac{n}{2}\right\rfloor\right)(n-2) + 1\).

What you did is you proved that $answer \leq \frac{1}{2} \left\lfloor \frac{n}{2}\right\rfloor \left(n - \left\lfloor \frac{n}{2}\right\rfloor\right)(n-2) + 1=M$ by providing an example of a fine subset(any $M$ triples would work). But where is the proof it is minimal? Moreover, think logically: if $c$ works for $n$, then it obviously works for $n+1$! So the answer is clearly monotonicly descending