If a triangle has two colours we can find exactly two vertices in this triangle with edges of different colours. Therefore if we call $a_i$ the number of red edges departing from vertex $i$, we find that the required number $N$ satisfies: \[N\ge {25\choose 3}-\frac{1}{2}\sum_{i=1}^{25}{a_i(24-a_i)}\ge {25\choose 3}-\sum_{i=1}^{25}{12^2}=500\] The approach is exactly the same as in here