We can divide the $$99$$ numbers into $$3$$ groups according to their remainder when divided by $$3$$. $$1$$, $$4$$, $$7\_\cdots$$ they belong to the group $$A$$, since they leave a remainder of $$1$$ when divided by $$3$$. $$2$$, $$5$$, $$8\_\cdots$$ they belong to the group $$B$$, since they leave a remainder of $$2$$ when divided by $$3$$. $$3$$, $$6$$, $$9\_\cdots$$ they belong to the group $$C$$, since they leave a remainder of $$0$$ when divided by $$3$$. As we know, the members in group $$C$$ cannot get together. Also, when the member of group $$A$$ and the member of group $$B$$ get together, they will create the sum which is a multiple of $$3$$. So the $$33$$ members of group $$C$$ will be in $$33$$ groups. Group $$A$$ and group $$B$$ can join in the $$33$$ groups however they want. 
The natural numbers from $$1$$ to $$99$$ inclusive are divided into n groups such that the following conditions hold:  Each number belongs to exactly one group.  Each group contains at least two numbers.  If two numbers belong to the same group, then their sum is not divisible by $$3$$.  What is the smallest number $$n$$ which satisfies the above conditions? ． 

A) $$3$$ 
B) $$33$$ 
C) $$9$$ 
D) $$66$$ 