We define $d_t(G)$ to be the minimum number of monochromatic edges of a t-coloring of $G$.
For example, $d_2(K_2)=0, d_2(K_3)=1, d_2(K_4)=2$ and in general, to calculate $d_2(K_n)$, we argue as follow: So let the colors be red and blue. If $n-1$ vertices are colored red, then those red colored vertices form a complete graph on $n-1$ vertices, and we need to remove $\binom{n}{2}$ edges. If $n-2$ vertices are colored red, then we need to remove $\binom{n-2}{2}+\binom{2}{2}$ edges. So If there are $k$ red colored vertices, then we need to remove $\binom{n-k}{2}+\binom{k}{2}$ edges.
So in general, $d_m(K_n)= min_{\text{all partition $\theta$ of n into sum of m positive numbers}}\sum_{j}\binom{\theta(j)}{2}$.
Please help give a tight upper bound.