Color integers of finite set $M\subseteq \mathbb{N}$ so that if different $a,b$ are the same color then $a+b$ is different color. $|M|_{\max}=?$ 
The integers $1, 2, \ldots, n$ are each colored with one of three colors, following the rule that if $a$, $b$ are distinct integers that both have the same color, $a+b$ must not be that color. What is the maximum possible value of $n$ for which this can be done?


Here is an example for $n=23$ (I can't find bigger $n$). 
$$A=\{9,10,11,12,13,14,15,16,17,18\}$$
$$B=\{3,5,6,7,19,20,21\}$$
$$C=\{1,2,4,8,22\}$$
But is this maximum value?

An unsatisfactory method, but:
By exhaustive search, there are only three solutions for $n=23$ :
$$\begin{align}
A &= \{1,2,4,8,11,22\},\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,16,17,18,20\}\end{align}$$
$$\begin{align}
A &= \{1,2,4,8,11,16,22\}\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,17,18,20\}\end{align}$$
$$\begin{align}
A &= \{1,2,4,8,11,17,22\}\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,16,18,20\}\end{align}$$
(Remarkably, they all share the same set $B$!)
None of these can be extended to a solution for $n=24$ (in all three cases, appending $24$ fails because $2+22=3+21=9+15=24$).
