An inequality on sequences with each term dividing sum of two neighbouring terms Positive integers $x_1, x_2, \dots, x_n$ ($n \ge 4$) are arranged in a circle such that each $x_i$ divides the sum of the neighbors; that is  $$\frac{x_{i-1}+x_{i+1}}{x_i} = k_i $$ is an integer for each $i$, where $x_0 = x_n$, $x_{n+1} = x_1$.  Prove that $$ 2n \le k_1 + k_2 + \dots + k_n < 3n. $$
This is a problem from some olympiad but I don't remember. 
What I did was like this :

Let $x_1$ be the smallest. Then obviously, $k_1\ge 2$. I tried to proceed inductively, but then I have no idea. I know it is not something useful, but I lack in any kind of idea. If I try induction, three new terms come replacing three previous terms. Which seems complicated and I can't do anything with it.

Sorry for such a bad try, but I seriously don't have any idea about it. Maybe something ingenious is required here. I hope someone can help. Thanks. 
EDIT : Removed meaningless ramblings of mine. Apologies.

I'm going to reformulate the problem in a slightly more convenient way (I rename $k_i-2$ as $k_i$).

Consider a sequence $\{(x_i,k_i)\}_{i=1}^n\subset \mathbb{R}^+\times
\mathbb{Z}$ such that $$ x_ik_i=(x_{i+1}-x_i)-(x_{i}-x_{i-1}). $$
  Prove that $0\leq \sum_{i=1}^nk_i<n.$

Upper Bound:
Since $x_i(k_i+2)=x_{i+1}+x_{i-1}>0$, each $k_i\geq -1$. Also notice that $\sum_{i=1}^n x_ik_i=0$. If $k_i=0$ for all $i$, we are done. Thus we may assume $k_1=-1$ without loss of generality.
Induct on $n$, starting at $n=2$. In this case we have
$$
 \{(2x,-1),(x,2)\},\qquad \mbox{ and }\sum_{i=1}^2 k_i=1<2.
$$
Now consider the sequence $\{(x_i,k_i)\}_{i=1}^n$ where $k_1=-1$. Then $x_1=x_2+x_n$; substituting into the $i=2$ and $i=n$ equations yields the sequence
$$
 \{(x_i',k_i')\}_{i'=1}^{n-1}=\{(x_2,k_2-1),(x_3,k_3),\cdots,(x_{n-1},k_{n-1}),(x_n,k_n-1)\}.
$$
Observe that the new sequence is still valid. Applying the inductive hypothesis,
$$
 \sum_{i=1}^n k_n = (-1) + \sum_{i=2}^n k_n = 1 + \sum_{i=1}^{n-1} k_n' < 1 + (n-1) = n.
$$
Lower Bound:
Apply AM-GM as follows.
$$
\sum_{i=1}^nk_i=\sum_{i=1}^n\left(\frac{x_i}{x_{i+1}}+\frac{x_{i+1}}{x_{i}}-2\right)\geq 0.
$$
