We extend periodically the definition $x_i$ to $\mathbb Z$. Let us write $S_O\doteqdot\sum_{i=1}^n x_{2i-1}$ and $S_E\doteqdot\sum_{i=1}^n x_{2i}$. Hence $S_O+S_E=4$. Without loss of generality we have $S_E\ge S_O$. Hence $S_E\ge 2$ so $S_E\ne 0$. Consider a function $f\colon\mathbb Z\to\mathbb R$ by $f(i)\doteqdot S_E\cdot x_{2i-1}-S_O\cdot x_{2i}$. It follows that $f$ has period $n$ and $$\sum_{i=1}^n f(i)=\sum_{i=1}^n(S_E\cdot x_{2i-1}-S_O\cdot x_{2i})=S_ES_O-S_OS_E=0.$$By gas station problem there exists a positive integer $a$ such that $\sum_{i=1}^b f(a+i-1)\ge 0$ for all nonnegative integers $b$, where the empty sum equals $0$. Without loss of generality assume $a=1$. Hence $\sum_{i=1}^b f(i)\ge 0$ for all nonnegative integers $b$. Let $p\doteqdot 1$ and let $q$ be the maximal nonnegative integer $b$ for which $$\sum_{i=1}^b (x_{2i-1}+x_{2i})\le 1+\frac{S_O}{S_E}.$$We will prove that $q$ exists, $q\le n-1$ and the ordered pair $(p,q)$ satisfies the problem requirements. First note that $$\sum_{i=1}^n (x_{2i-1}+x_{2i})=4>2\ge 1+\frac{S_O}{S_E}.$$Since all $x_1,x_2,\ldots,x_{2n}$ are all nonnegative then $q$ exists and $q\le n-1$. It remains to prove that $$\sum_{i=1}^q x_{2i} \le 1 \mbox{ and } \sum_{i=q+1}^{n-1} x_{2i+1} \le 1$$or equivalently $$\sum_{i=1}^q x_{2i}\le 1 \mbox{ and } \sum_{i=q+2}^n x_{2i-1} \le 1.$$First note that \begin{align*} (S_O+S_E)\sum_{i=1}^q x_{2i}&=\sum_{i=1}^q(S_Ox_{2i}+S_Ex_{2i})\\ &=\sum_{i=1}^q(S_Ex_{2i-1}-f(i)+S_Ex_{2i})\\ &=S_E\sum_{i=1}^q(x_{2i-1}+x_{2i})-\sum_{i=1}^q f(i)\\ &\le S_E\left(1+\frac{S_O}{S_E}\right)\\ &=S_O+S_E. \end{align*}Hence $\sum_{i=1}^q x_{2i}\le 1$. It remains to prove that $\sum_{i=q+2}^n x_{2i-1}\le 1$. By maximality of $q$ we obtain $$\sum_{i=1}^{q+1} (x_{2i-1}+x_{2i})\ge 1+\frac{S_O}{S_E}.$$Therefore $$\sum_{i=q+2}^n (x_{2i-1}+x_{2i})\le 3-\frac{S_O}{S_E}.$$Moreover we know that $\sum_{i=1}^{q+1} f(i)\ge 0$ and $\sum_{i=1}^n f(i)=0$, hence $\sum_{i=q+2}^n f(i)\le 0$. It follows that \begin{align*} (S_O+S_E)\sum_{i=1}^q x_{2i-1}&=\sum_{i=1}^q(S_Ox_{2i-1}+S_Ex_{2i-1})\\ &=\sum_{i=1}^q(S_Ox_{2i-1}+S_Ox_{2i}+f(i))\\ &=S_O\sum_{i=1}^q(x_{2i-1}+x_{2i})+\sum_{i=1}^q f(i)\\ &\le S_O\left(3-\frac{S_O}{S_E}\right). \end{align*}Therefore \begin{align*} \sum_{i=1}^q x_{2i-1}&\le\frac{S_O}{S_O+S_E}\left(3-\frac{S_O}{S_E}\right)\\ &=\frac{S_O}{4}\left(3-\frac{S_O}{4-S_O}\right)\\ &=\frac{S_O}{4}\cdot\frac{12-4S_O}{4-S_O}\\ &=\frac{S_O(3-S_O)}{4-S_O}\\ &=1-\frac{(S_O-2)^2}{4-S_O}\\ &\le 1 \end{align*}as desired.

Cycle Digraph and EV for the win. Wait, remember the car gas joke problem? For convenience, have the numbers on the circle be $a_1, b_1, a_2, b_2, \cdots, a_n, b_n$ (indices for $a,b$ are mod $n$) Assume for contradiction for every $p$, there doesn't exist $q$ such that $\sum\limits_{i=1}^q x_{p+2i-1} \le 1$ and $\sum\limits_{i=q+1}^{n-1} x_{p+2i}\le 1$. Let $q$ be the maximal $q$ such that $\sum\limits_{i=1}^q x_{p+2i-1} \le 1$. This means for every $p$, there is a $f(p)=q+1$ such that $\sum\limits_{i=f(p)}^{n-1} x_{p+2i}> 1$ and $\sum\limits_{i=1}^{f(p)} x_{p+2i-1} > 1$ Suppose $x_{p+1}=a_k, x_{p+2}=b_k$, then $\sum\limits_{i=1}^{f(p)} x_{p+2i-1} = \sum\limits_{i=0}^{f(p)-1} a_{k+i} $ and $\sum\limits_{i=f(p)}^{n-1} x_{p+2i} = \sum\limits_{i=f(p)-1}^{n-2} b_{k+i}$ For every single $k$, let $g(k)$ (k maps to $p$) satisfy $\sum\limits_{i=k}^{g(k)-1} a_i > 1$ and $\sum\limits_{i=g(k)-1}^{k+n-2} a_i > 1$. Consider the directed graph with vertices in $\{1,\cdots,n\}$ and an edge $k\to g(k)$. It has a cycle, call $n_1\to n_2\to \cdots \to n_k \to n_1$ Summing, $$a_{n_1}+\cdots+a_{n_2-1} > 1,$$ $$a_{n_2}+ \cdots + a_{n_3-1}>1$$ All the way to $$a_{n_k}+\cdots+a_{n_k-1}>1$$ Consider a pointer that starts from $n_1$ that goes clockwise to $n_2$, $n_3$ and eventually goes back to $n_1$. Suppose it completes $x$ revolutions. This implies $x(a_1+\cdots+a_n)=\sum\limits_{j=1}^k \sum\limits_{i=n_j}^{n_{j+1}-1} a_i > k$ so $a_1+\cdots+a_n > \frac kx$. Similarly, $b_1+\cdots+b_n>\frac{k}{k-x}$. Adding the two inequalities give $4=a_1+\cdots+a_n+b_1+\cdots+b_n>\frac kx + \frac{k}{k-x}\ge 4$, absurd!

Let $q$ be the maximal $q$ such that $\sum\limits_{i=1}^q x_{p+2i-1} \le 1$. This means for every $p$, there is a $f(p)=q+1$ such that $\sum\limits_{i=f(p)}^{n-1} x_{p+2i}> 1$ and $\sum\limits_{i=1}^{f(p)} x_{p+2i-1} > 1$

Let us write $[n]\doteqdot\{1,2,\ldots,n\}$ for brevity. Lemma. (gas station lemma) Let $n$ be a positive integer and let $a_1,\ldots,a_n$ be real numbers such that $\sum_{i=1}^n a_i=0$. Then there exists an integer $p$ such that $\sum_{i=1}^k a_{p+i-1}\ge 0$ for all positive integers $k\in[n]$, where the indices are taken modulo $n$. ProofPick an integer $c$ such that $0\le c\le n-1$ and $\sum_{i=1}^c a_i$ is minimal, where the empty sum is $0$. We claim that $p\doteqdot c+1$ works. To prove this, let $k\in[n]$ be a positive integer. We wish to prove that $\sum_{i=1}^k a_{p+i-1}\ge 0$. Equivalently: \begin{align*} \iff&\sum_{i=c+1}^{c+k} a_{i}\ge 0\\ \iff&\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i. \end{align*}We distinguish between two cases. Case 1: $c+k<n$. By minimality we obtain $\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i$. Case 2: $c+k\ge n$. Note that $c+k\le (n-1)+n\le 2n-1$. Hence $0\le c+k-n\le n-1$. Since $\sum_{i=1}^n a_i=0$ and by minimality we obtain $$\sum_{i=1}^{c+k}a_i=\sum_{i=1}^{c+k-n}a_i\ge\sum_{i=1}^{c}a_i.$$ Since we have exhausted all cases, the lemma is proved. Now define real numbers $a_1,b_1,a_2,b_2,\ldots,a_n,b_n$ by $a_i\doteqdot x_{2i-1}$ and $b_i\doteqdot x_{2i}$ for all $i\in[n]$. It suffices to prove that there exist integers $p$ and $q$ such that $0\le q\le n-1$ and $$\sum_{i=q+2}^{n} a_{p+i}\le 1\mbox{ and }\sum_{i=1}^q b_{p+i}\le 1,$$where indices are taken modulo $n$. The assignment $(p,q)\mapsto (2p+1,q)$ solves the original problem. Define $A\doteqdot\sum_{i=1}^n a_i$ and $B\doteqdot\sum_{i=1}^n b_i$. If $A\le 1$ then we can choose $(p,q)=(0,0)$. Similarly, if $B\le 1$ then we can choose $(p,q)=(0,n-1)$. So assume $A>1$ and $B>1$. Define real numbers $c_1,\ldots,c_n$ by $c_i\doteqdot Ba_i-Ab_i$ for all $i\in[n]$. Note that $$\sum_{i=1}^n c_i=B\sum_{i=1}^n a_i-A\sum_{i=1}^n b_i=BA-AB=0.$$By the gas station lemma, there exists an integer $p$ such that $\sum_{i=1}^k c_{p+i}\ge 0$ for all integers $k\in[n]$, where indices are taken modulo $n$. Let $q$ be the maximal integer $q$ such that $0\le q\le n$ and $\sum_{i=1}^q b_{p+i}\le 1$. Since we assume $B>1$ then $q\le n-1$. Assume by contradiction that $\sum_{i=q+2}^n a_{p+i}>1$. By maximality we obtain $\sum_{i=1}^{q+1} b_{p+i}>1$. Hence $$0\le\sum_{i=1}^{q+1}c_{p+i}=B\sum_{i=1}^{q+1} a_{p+i}-A\sum_{i=1}^{q+1} b_{p+i}<B\sum_{i=1}^{q+1} a_{p+i}-A.$$Therefore $\sum_{i=1}^{q+1} a_{p+i}>\frac{A}{B}$. Now we have $$\sum_{i=q+2}^n c_{p+i}=\sum_{i=1}^n c_{p+i}-\sum_{i=1}^{q+1} c_{p+i}=-\sum_{i=1}^{q+1} c_{p+i}\le 0.$$Hence $$0\ge\sum_{i=q+2}^n c_{p+i}=B\sum_{i=q+2}^n a_{p+i}-A\sum_{i=q+2}^n b_{p+i}>B-A\sum_{i=q+2}^n b_{p+i}.$$Therefore $\sum_{i=q+2}^n b_{p+i}>\frac{B}{A}$. Now by AM-GM \begin{align*}\sum_{i=1}^n(a_{p+i}+b_{p+i})&=\sum_{i=1}^{q+1} a_{p+i}+\sum_{i=1}^{q+1} b_{p+i}+\sum_{i=q+2}^n a_{p+i}+\sum_{i=q+2}^n b_{p+i}\\ &>\frac{A}{B}+1+1+\frac{B}{A}\\ &>4, \end{align*}contradiction.

For each $i\in[n]$, define $a_i\doteqdot x_{2i-1}$ and $b_i\doteqdot x_{2i}$. Define $A\doteqdot\sum_{i=1}^n a_i$ and $B\doteqdot\sum_{i=1}^n b_i$. Hence $A+B=4$. WLOG assume $B\ge 2$. Define real numbers $c_1,\ldots,c_n$ by $c_i\doteqdot Ba_i-Ab_i$ for all $i\in[n]$. Note that $$\sum_{i=1}^n c_i=B\sum_{i=1}^n a_i-A\sum_{i=1}^n b_i=BA-AB=0.$$By the gas station lemma, there exists an integer $r$ such that $\sum_{i=1}^{k} c_{r+i-1}\ge 0$ for all integers $k\in[n]$, where indices are taken modulo $n$. WLOG assume $r=1$ for simplicity, so $\sum_{i=1}^{k} c_{r}\ge 0$ for all integers $k\in[n]$. Now we define $p=1$ and $q$ is the maximal nonnegative integer such that $\sum_{i=1}^{q} b_i\le 1$. We will prove that the pair $(p,q)$ satisfies the conditions of the problem statement. Since $B\ge 2$ we have $q\le n-1$. Moreover, we have $$\sum_{i=1}^q x_{p+2i-1}=\sum_{i=1}^q x_{2i}=\sum_{i=1}^q b_i\le 1.$$Moreover, observe that $$\sum_{i=q+1}^{n-1} x_{p+2i}=\sum_{i=q+1}^{n-1} x_{1+2i}=\sum_{i=q+2}^{n} x_{2i-1}=\sum_{i=q+2}^{n} a_i.$$Hence, it remains to prove that $\sum_{i=q+2}^{n} a_i\le 1$. Observe that $$0\le\sum_{i=1}^{q+1} c_i=B\sum_{i=1}^{q+1} a_i-A\sum_{i=1}^{q+1} b_i\le B\sum_{i=1}^{q+1} a_i-A.$$Hence $\sum_{i=1}^{q+1} a_i\ge\frac{A}{B}.$ As a result, $\sum_{i=q+2}^{n}a_i\le A-\frac{A}{B}$. Hence, it suffices to prove that $A-\frac{A}{B}\le 1$, or equivalently $AB\le A+B$. The last inequality follows from $A+B=4$ and AM-GM: $$AB\le\left(\frac{A+B}{2}\right)^2=4=A+B.$$