The answer is $89$. An example is: $C_1 = \{1\}, C_2 = \{2\}, C_3 = \{4\} C_4 = \{3,5\}, C_5 = \{8\}, C_6 = \{6,10\}, C_7 = \{7,9\} $. Here we've designated all numbers less or equal than $10$. $C_8 = \{16\}, C_{9} = \{32\}, C_j = \{(j+1), 32-(j+1)\}$ for $j = 10,11,12,13,14$. Until now we've designated all numbers less than $22$, and also $32$ $C_{15} = \{64\}, C_{l} = \{l+6, 64 - (l+6)\}$ for $l = 16, 17, \cdot \cdot \cdot, 25$. Until now we've designated all numbers less or equal than $42$, and also $64$. $C_{26} = \{128\}, C_i = \{i + 16, 128 - (i+16)\}$, for $i= 27, 28, \cdot \cdot \cdot , 47$. Until now we've designated all numbers less or equal than $85$, and also $128$. Now, take $C_m = \{m+38, 256 - (m+38)\}$ for $m = 48, 49, \cdot \cdot \cdot , 89$, and finally we've designated all numbers of the set $\{1,2, \cdot \cdot \cdot 170\}$. Let's prove that this is the largest possible value of $k$. This is simple! Note that the maximum possible value of numbers that are alone in their own colors is $8$. In other case, we would have $a_1 \mid a_2 \mid \cdot \cdot \cdot \mid a_9 \Rightarrow 170 \geq a_9 \geq 2^{8}a_1 \geq 256$, with $1 \leq a_1 \neq \cdot \cdot \cdot \neq a_9 \leq 170$. Since in our example we have $8$ numbers alone in their own colors and all the other colors have exactly $2$ numbers on it, we are maximizing the number of colors. So we are done! $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \blacksquare$

your example is wrong

hectorleo123 wrote: your example is wrong yeah it was a typo. fixed up.