Hrvatska 2011

Assign numbers $1,2,\dots ,2n$ to the vertices of the observed $2n$-gon respectively. Let $a_k$ be the number assigned to the vertex $P_k$. Then $a_1,a_2,\dots ,a_{2n}$ is a permutation of $1,2,\dots ,2n$.



Segments $\overline{P_i{P}_{i+1}}$ and $\overline{P_j{P}_{j+1}}$ for $i\ne j$ are parallel if and only if they determine an isosceles trapezoid with bases $\overline{P_i{P}_{i+1}}$ and $\overline{P_j{P}_{j+1}}$. Its legs (or diagonals) $\overline{P_i{P}_{j+1}}$ and $\overline{P_j{P}_{i+1}}$ are congruent and a rotation around the circumcenter maps points $P_i$ and $P_j$ into points $P_{j+1}$ if $P_{j+1}$ are mapped respectively.

That is why the condition $\overline{P_i{P}_{i+1}}\parallel \overline{P_j{P}_{j+1}}$ is equivalent to the condition that the number of vertices between $P_i$ and $P_{j+1}$ is equal to the number of vertices between $P_j$ and $P_{i+1}$, taking into account the orientation. That condition can be written as $a_i-a_{j+1}\equiv a_j-a_{i+1}(\mathrm{mod}2n)$, i.e. $$a_i+a_{i+1}\equiv a_j+a_{j+1}(\mathrm{mod}2n).$$ Now assume that none of the given segments are parallel. Then the sums $a_k+a_{k+1}$ give different remainders modulo $2n$, so $$\sum _{k=1}^{2n}(a_k+a_{k+1})\equiv 0+1+\cdots +(2n-1)(\mathrm{mod}2n)$$ The sum on the right hand side is $n(2n-1)$ which is congruent to $n$ modulo $2n$.

On the other hand, we have $$\sum _{k=1}^{2n}(a_k+a_{k+1})=2\sum _{k=1}^{2n}{a}_k=2\sum _{k=1}^{2n}k=2n(2n+1)$$ The resulting expression is divisible by $2n$, which is in contradiction with the conclusion above. Therefore our assumption was incorrect which means there is at least one pair of parallel segments.