This is too bad to write up - especially as there is no good way to convert my TikZ diagrams to Asymptote to use on AoPS. Unfortunately, the solution in BWM 2017 on my website is in German but you can probably get a sense of the construction by looking at the diagrams. Answer. $1345$. Proof. By angle sum, it's easy to show that the answer must be at most $1345$. Let $w$ be the number of acute angles in the $2017$-gon, $S$ the sum of the acute angles and $\overline{S}$ the sum of the obtuse angles. Then \[ 360^{\circ} \cdot (2017-w) > \overline{S} = 2015 \cdot 180^{\circ}-S> 2015 \cdot 180^{\circ}-90^{\circ} \cdot w \]or $363420 > 270w$ or $1346>w$. The construction for such a $2017$-gon is the non-trivial part and mainly annoying. I will give a sketch. Take a regular $1342$-gon $E_1E_2 \dots E_{1342}$ and put "almost-squares" on the segments $E_1E_2,E_2E_3,\dots,E_{671}E_{672}$. By "almost-squares" I mean quadrilaterals with say angles $90.1^{\circ}$ and $89.9^{\circ}$. The larger angle should hit the $1342$-gon. (I want to refer to the diagrams in my solution on my website, as a picture is worth a thousand words - or in this case even more.) We will then remove the edges $E_1E_2,E_2E_3,\dots,E_{671}E_{672}$ and also most of the rest of the $1342$-gon. Instead we want to construct a "fitting" triangle with one side going through $E_1E_{672}$ and will somewhat have constructed a figure that works. That's a (very very) rough sketch. Again, diagrams would help. Feel free to ask for more details. (Perhaps I've also missed some super easy construction - pardon then, but please post the solution in this case!)

Link is not reachable.I can't see your solution on your website.

Olympus_mountaineer wrote: Link is not reachable.I can't see your solution on your website. No it is there, but not directly! You have to click on another link (scroll down, and) "BWM 2017, Round 1".