A similar question appeared last week.
In this problem, I want to split the integers from $1$ to $3n$ into triples $\{x,y,z\}$ with $x+y=4z$ instead of $3z$.
I have a set of solutions, but am looking for more ideas.
$\sum(x+y+z)=\sum5z$ so $n=5k$ or $5k+3$. There are already hundreds of solutions for $n=13$, and I imagine that number will only increase with $n$, but the trick is to find one for any particular $n$.
Any solution for $n=N$ can be extended to one for each of $n=19N-7,19N+8,19N+13$. That gives solutions for $3^m$ different $n$ below $Z=19^m$, which is more than $\sqrt[3]Z$ different $n$ below $Z$. But most solutions do not contain a smaller one within them.
To split the numbers from $3N+1$ to $57N-21$ into triples $$(3N+1+k,33N-9+3k,9N-2+k)$$ for $k =0..6N-4$ and $$(51N-18+k,33N-10+3k,21N-7+k)\\ (27N-9+k,33N-11+3k,15N-5+k)$$ for $k =0..6N-3$.
A way to split $\{3N+1,...,3(19N+8)\}$ into triples is $$(3N+1+k,33N+15+3k,9N+4+k)\\ (51N+22+k,33N+14+3k,21N+9+k)$$ for $k=0..6N+2$, and $$(27N+12+k,33N+16+3k,15N+7+k)$$ for $k=0..6N+1$
To split $\{3N+1 ...,3(19N+13)\}$ into triples $$(3N+1+k,33N+23+3k,9N+6+k)$$ for $k=0..6N+4$ and $$(27N+19+k,33N+25+3k,15N+11+k)\\ (51N+36+k,33N+24+3k,21N+15+k)$$ for $k=0..6N+3$.
At Split $\{1,...,3n\}$ into triples with $x+y=5z$ - no solutions?, Thomas Andrews has shown there are no solutions for $x+y=5z$.