All six digits of three 2-digit numbers are different. If $$N$$ is the largest possible sum of three such numbers, what is the sum of digits of $$N$$?

To maximise the sum, use the six largest digits $$4, 5, 6, 7, 8, 9$$, and place the three largest of them in the tens places, because swapping a larger digit from a units place into a tens place increases the sum by $$9(v - u) > 0$$. This gives $$N = 10(9 + 8 + 7) + (6 + 5 + 4) = 255$$, attained for instance by $$96 + 85 + 74$$. The digit sum of 255 is $$2 + 5 + 5 = 12$$.