, 2022

Proof: Set $A = x_1 + x_3 + \dots + x_{2n-1}$ and $B = x_2 + x_4 + \dots + x_{2n}$.
If one of $A, B$ is less than or equal to 1, the problem is obvious. If $A > 1$ and $B > 1$, for $0 \le k \le n-1$, let $m(k) \in \{1, 2, \dots, n-1\}$ be the unique integer such that
$$\sum_{i=0}^{m(k)} x_{2k+2i+1} > 1 \quad (1)$$
$$\sum_{i=0}^{m(k)-1} x_{2k+2i+1} \le 1. \quad (2)$$
Note that, if $x_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \dots + x_{2k+2n-2} \le 1$, then this equation and (2) gives what we needed with $p = 2k$ and $q = m(k)$.
Now suppose that
$$x_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \dots + x_{2k+2n-2} > 1. \quad (3)$$
Now, we construct an oriented graph with $0, 1, 2, \dots, n-1$ as vertices, and connect an edge from each $k$ to $k+m(k)+1$ (with indices modulo $n$). It is easy to see that there are loops in this graph. We may assume that $k_1 \to k_2 \to \dots \to k_t \to k_1$ is a minimal loop.
If $t=1$, then $\sum_{i=0}^{n-2} x_{2k_1+2i+1} \le 1$. Setting $p=2k_1$ and $q=n-1$ gives what we want.
If $t>1$, then set
$$\left\{ \frac{k_2 - k_1}{n} \right\} + \left\{ \frac{k_3 - k_2}{n} \right\} + \dots + \left\{ \frac{k_t - k_{t-1}}{n} \right\} + \left\{ \frac{k_1 - k_t}{n} \right\} = s.$$
In other words, when $k = k_1, k_2, \dots, k_t$, there are totally $s \cdot n$ elements in the sums of (1). Moreover, using the condition for defining the edges, we know that every term in $x_1, x_3, \dots, x_{2n-1}$ appeared in the sum (1) exactly $s$ times. On the other hand, when $k = k_1, k_2, \dots, k_t$, there are in total $(t-s) \cdot n$ elements in the sums in (3). The condition for linking an edge shows that each of $x_2, x_4, \dots, x_{2n}$ appeared exactly $t-s$ times in the above sum; this is because $x_{2j}$ appeared in (3) if and only if $x_{2j+1}$ did not appear in (1).
Now take the total sum of (1) when $k = k_1, k_2, \dots, k_t$, we get $s \cdot A > t$. Similarly, taking the total sum of (3) when $k = k_1, k_2, \dots, k_t$ gives $(t-s) \cdot B > t$. But $A + B > t(\frac{1}{s} + \frac{1}{t-s}) \ge 4$. This leads to a contradiction. The problem is proved.