Hmm... I can't prove it for $n=1$ (Kuba said me that it is the hardest thing in this problem ) All over cases can be derived from $n=1$. Let's show it. Suppose the contrary, i.e. there are no triangle of area $\leq 1/6n$. Consider $4n+2$-gon with $n\geq 2$ and cut a 6-gon from it by mean of a diagonal ($AB$ on figure). Let's prove that the triangle of area $<1/6n$ can't adjoin the diagonal $AB$. Otherwise if $ABC$ is a such triangle then draw a line throw $A$ parallel to $BC$. We will necessary have $D$ and $E$ lies in the other halfplane than $BC$. So $4n+2$ is not convex.Therefore we can divide $4n+2$-gon into $n$ 6-gons by mean of diagonals and no triangle which adjoins these diagonals will have area $<1/6n$.

Myth wrote: Hmm... I can't prove it for n=1 (Kuba said me that it is the hardest thing in this problem )

Nice problem, by the way!

Nice idea, Myth! I can prove the n=1 case: In hexagon ABCDEF, $P=AD\cap CF$, $Q=CF\cap EB$, $R=EB\cap AD$. The six triangles APB, BPC, CQD, DQE, ERF, FRA cover less than the whole hexagon (if the three diagonals are not concurrent) or the whole triangle (if they are). In either case, one of those six triangles will have area $\le \frac{1}{6}(\text{area})$, so say it is APB. Then one of AFB and ACB has area at most the area of APB, so we're done.

Sorry, I don't see it. Indeed, the case described by Fedor is clear. But common case... I am missing something.

I really don't understand what Myth did, especially the part: Myth wrote: We will necessary have $D$ and $E$ lies in the other halfplane than $BC$. And also, how exactly is the case $n>1$ derived from $n=1$?

If $E$ lies in the same halfplane, then $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction. The general cases can be done in the following way: cut the polygon into $n$ hexagons by mean of diagonals. One of this hexagon have area $\leq 1/n$ and we can cut a tringle of area $\leq 1/6$ from it, therefore this triangle is at most $\frac{1}{6n}$ of the area of the initial polygon. And we prooved that this triangle can't be "inner" triangle, so it is a required triangle.

okay. now i understand the general case. Myth wrote: If $E$ lies in the same halfplane, then $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction. i still don't understand the first part. Why $S_{EBC}\leq S_{ABC}\leq \frac{1}{6n}$ yields a contradiction?

Because we supposed the contrary, i.e. there is no such triangle.

Case n=1 is also here