my solution: hint:(a-b)(c-b)(a-d)(c-d) in detail: product (a-b)(c-b)(a-d)(c-d) for every $e_i,e_j, 2|i+j,i<j$ a,c be the number on $e_j$, b,d be the number on $e_i$ for point x,y if they are not ajjectent then in the product, it will occur twice, if it's ajjectent it's appears only once, and also chick the point that are on $e_1$ and $e_n$ then we get it.

A different solution than the above is to let $d_1 < d_2 < \cdots < d_n$ be a permutation of the $e_i$'s, and consider the effect of switching the labels of the vertices labelled $d_i$ and $d_{i+1}$. By casework (on whether the two vertices are adjacent or not), one can easily check that the difference of the number of alternating pairs of sides and the number of positive sides is always of the same parity. Now, it suffices to simply label the edges in increasing order in a clockwise direction and check that this quantity is always odd.

Consider the operation defined by increasing a single number to make it the maximum number. We first show that such an operation changes the parity of the number of alternating pairs of sides and the number of positive sides in the same way: Label the vertices clockwise $v_1,v_2,\ldots, v_n$ and WLOG we operate on $v_2$. Note that if the initial value of $v_2$ was between that of $v_i$ and $v_{i+1}$, then increasing it will flip the alternating nature of the pair $v_iv_{i+1},v_{i\pmod 2}v_{i\pmod 2+1}$. On the other hand, if the initial value of $v_2$ was not between them, then the operation preserves alternating nature. So, the number of alternating pairs of sides changes by the number of edges for which the initial value of $v_2$ was between. This is odd if $v_2$ is between $v_1$ and $v_3$ and even otherwise. However, the parity of the change in the number of positive sides is odd if $v_2$ was between $v_1,v_3$ and even otherwise, so we have the desired claim. Now, as we can get to any relative ordering with the above operation starting from a sorted state, it suffices to prove the result for the sorted state, which is trivial as there are $0$ alternating pairs, and an even number of positive sides.

We say that a side $e_i$ is even if $i$ is even, and define odd sides similarly. We construct a graph whose vertices are the even sides and two vertices are adjacent if and only if they form an alternating pair, similarly construct the graph for odd sides. We now swap the numbers on consecutive vertices $V_i$ and $V_{i+1}$. WLOG assume $V_iV_{i+1}$ is even, then the even graph is still the same. For the odd graph, suppose the numbers on vertex $V_k$ is $v_k$, then in the graph, $(v_{i-1},v_i), (v_{i+1},v_{i+2})$ becomes $(v_{i-1},v_{i+1}),(v_i,v_{i+2})$. The main observation is that for another vertex $(x,y)$, then it forms an alternating pair with $(v_{i-1},v_i)$ if and only if $$(x-v_{i-1})(x-v_i)(y-v_{i-1})(y-v_i)<0$$Therefore, the parity of the number of edges between $(x,y)$ and $(v_i,v_{i-1}),(v_{i+1},v_{i+2})$ is the sign of $$\prod_{i-1\leq k\leq i+2}(x-k)(y-k)$$which remains invariant after the transposition. Therefore, the only change on the sign is the edge between the two vertices, which is $$\frac{(v_{i-1}-v_{i+1})(v_{i-1}-v_{i+2})(v_i-v_{i+1})(v_i-v_{i+2})}{(v_{i-1}-v_i)(v_{i-1}-v_{i+2})(v_{i+1}-v_i)(v_{i+1}-v_{i+2})}=-\frac{(v_{i-1}-v_{i+1})(v_i-v_{i+2})}{(v_{i-1}-v_i)(v_{i+1}-v_{i+2})}$$