, 2011

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

Segments $\overline{P_iP_{i+1}}$ and $\overline{P_jP_{j+1}}$ for $i \neq j$ are parallel if and only if they determine an isosceles trapezoid with bases $\overline{P_iP_{i+1}}$ and $\overline{P_jP_{j+1}}$. Its legs (or diagonals) $\overline{P_iP_{j+1}}$ and $\overline{P_jP_{i+1}}$ 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 $\overline{P_iP_{i+1}} \parallel \overline{P_jP_{j+1}}$ 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 $a_i - a_{j+1} \equiv a_j - a_{i+1} \pmod{2n}$, i.e.
$$a_i + a_{i+1} \equiv a_j + a_{j+1} \pmod{2n}.$$
Now assume that none of the given segments are parallel.
Then the sums $a_k + a_{k+1}$ give different remainders modulo $2n$, so
$$\sum_{k=1}^{2n} (a_k + a_{k+1}) \equiv 0 + 1 + \dots + (2n - 1) \pmod{2n}$$
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.