I too cant understand the editorial.During the contest I wasted a lot of time on this problem trying to model it into some sort of matching problem.Does a matching solution exist for this problem?

why that upperbound is achievable

The simplest and the best proof of this is to show by example.

how to come up with a formula

Actually, there are a lot of formulas. I can show how i came up with my solution.

At first, I decided to take first set as 0 1 2 ... k-1.

Secondly, I started to think about the difference between adjacent elements, sum of triple such differences should be equal to zero. And after a little reasoning, i decided to take second set of differences as -(x+1), +x, -(x+1), +x..., so the third set will be +x, -(x+1), +x, -(x+1), where . After that, i had only to adjust the coefficients for x and the first element of the second set.

Also, there is a very simple solution with a cyclic shift. It looks like:

0 1 2 3 4 5 6 7 8
4 5 6 7 8 0 1 2 3
8 6 4 2 0 7 5 3 1

#include <iostream>
using namespace std;
int main(){
	freopen("in","r",stdin);
	int n; cin>>n;
	int k = (2*n)/3;
	cout<<k+1<<endl;
	int y = 2*k-n;
	int x = n-2*y;
	for(int  i=0;i<=y;i++){
		cout<<i<<" "<<x+i<<" "<<n-x-2*i<<endl;
	}
	for(int  i=0;i<k-y;i++){
		cout<<y+i+1<<" "<<i<<" "<<n-y-1-2*i<<endl;
	}
	return 0;
}

This is why it is achievable