Prove there is such a triple There are $3n$ colored numbers from $1$ to $3n$ such that each color is represented exactly by $n$ numbers. Prove there is 3-colored triple $a,b,c$ such as $$a+b=c$$
I started by assuming various values for minimum each of color, but didnt come to any conclusions(

Note that this problem was initially asked and solved by Alekseev and Savchev, in the Kvant journal, 4:23, problem M1040.
Let $A,B,C$ be the three monochromatic subsets of $[3n]$. Without loss of generality let $1,\ldots,k-1$ be in $A$ (i.e. the first $k-1$ integers are in $A$, with $k-1\geq1$), and let $k\in B$.
We call three numbers a good triple if they satisfy your condition. Suppose that there are no good triple. 
Let $a\in C$ be any number. Note that $a-1\not\in B$ as otherwise $(1,a-1,a)$ would be a good triple. 
Suppose that $a-1\in C$, and consider 


*

*The integer $a-k$. If it is in $A$ then $(a-k,k,a)$ would be a good triple. If it is in $B$, then $(k-1,a-k,a-1)$  would be a good triple. Therefore $a-k \in C$.

*The integer $a-k-1$. If it is in $A$ then $(a-k-1,k,a-1)$ would be a good triple. If it is in $B$, then $(1,a-k-1,a-k)$  would be a good triple. Therefore $a-k-1 \in C$.

*The integer $a-2k$. If it is in $A$ then $(a-2k,k,a-k)$ would be a good triple. If it is in $B$, then $(k-1,a-2k,a-k-1)$  would be a good triple. Therefore $a-k \in C$.

*...


Repeating this argument we conclude that all integers of the form $a-ik$ and $a-ik-1$ ($i=0,1,\ldots$) are in $C$. But note that because $a>k$, there exist some $i$ such that $ik\leq a \leq i(k+1)$ and therefore such that $1\leq a-ik \leq k$. And we know that this number is either in $B$ (if it is $k$) or in $A$ (if is it $<k$). Hence a contradiction, and $$a-1\in A$$
Therefore we proved that $\forall a\in C, \ a-1 \in A$. But note that $k-1\in A$ while $k\in B$ therefore there is at least one element $a\in A$ for which $a+1\not\in C$. Therefore $\vert A\vert > \vert C\vert$, a contradiction, and there must be at least one good triple.
