We number the arcs with $0, 1, 2, \dots, 2n - 1$, clockwise and assume that the arcs with numbers $a_i$ and $b_i$ have $i$ written on them. By the problem's assumption
$$a_i + b_i \equiv a_i - b_i \equiv \pm(i + 1) \equiv i + 1 \pmod{2}.$$
Then by adding up these equalities modulo $2$ we have
$$\begin{align*} \sum_{i=0}^{n-1} (i+1) &\equiv \sum_{i=0}^{2n-2} (a_i + b_i) = \sum_{i=0}^{2n-1} i \\ &\implies \frac{n(n+1)}{2} \equiv n(2n-1). \end{align*}$$
The number $1399$ does not satisfy the last equation and this concludes the proof. ■