Min-max To The Max!

I did it the way you did an earlier problem.

Let's assume $1\leq x,y,z\leq 100$ throughout, so that there are $100^3=1000000$ summands. Now $\sum\max(x,y,z)+\sum\min(x,y,z)$ $=\sum\max(x,y,z)+\sum(101-\max(101-x,101-y,101-z))$ $=\sum\max(x,y,z)+\sum(101-\max(x,y,z))$ $=\sum101$ $=101(100^3)$ $=\boxed{101000000}$

Small Typo in the problem: They are ordered triples of integers, of course.

Let $\text{med}(x,y,z)$ stand for the median of the three numbers. Then $\text{max}(x,y,z) + \text{min}(x,y,z) = (x + y + z) - \text{med}(x,y,z).$ Now $\sum_{1\leq x,y,z\leq n} x = n^2 \sum_{1\leq x\leq n} x = \tfrac12 n^3(n+1),$ and likewise for $y, z$ ; so that $\sum_{1\leq x,y,z\leq n} (x + y + z) = \tfrac32 n^3(n+1).$ Now for the sum of the medians. We can pair each triple $(x, y, z)$ with $(n+1-x, n+1-y, n+1-z)$ ; if the median of the first triple is $a$ , then that of the second triple is $n+1-a$ . On average, then, each half of the pair contributes $\tfrac12(n+1)$ to the sum. (If $n$ is odd, the triple $(\tfrac12(n+1), \tfrac12(n+1), \tfrac12(n+1))$ remains unpaired, but this does not change our argument.) Therefore $\sum_{1\leq x,y,z\leq n} \text{med}(x,y,z) = \sum_{1\leq x,y,z\leq n} \tfrac12(n+1) = \tfrac12 n^3(n+1).$ Finally, $\sum \text{max}(x,y,z) + \text{min}(x,y,z) \\ = \sum (x + y + z) - \sum \text{med}(x,y,z) \\ = \tfrac32 n^3(n+1) - \tfrac12 n^3(n+1) = n^3(n+1).$ With $n=100$ we find the answer $\boxed{101\:000\:000}$ .