Solution: The left-hand inequality can be easily proved by A.M. $\ge$ G.M., for example:
$$\begin{aligned} \sum_{j=1}^{n} k_j &= \sum_{j=1}^{n} \left( \frac{x_{j-1}}{x_j} + \frac{x_j}{x_{j-1}} \right) \\ &\ge 2 \sum_{j=1}^{n} \sqrt{\frac{x_{j-1}}{x_j} \cdot \frac{x_j}{x_{j-1}}} = 2n. \end{aligned}$$

Below we deal with the right-hand inequality. In fact, the assumption $n \ge 4$ is not so important; we will prove it for all positive integers $n$. We denote the middle sum by $S_n$, and what we need to prove is $S_n < 3n$.

When $n=1$, the left and right neighbors of $x_1$ are itself. Therefore
$$S_1 = k_1 = \frac{x_1 + x_1}{x_1} = 2 < 3 = 3 \cdot 1.$$

When $n = 2$, from the given conditions we know $k_1 = \frac{x_2+x_2}{x_1}$, i.e., $k_1x_1 = 2x_2$. Similarly we get $k_2x_2 = 2x_1$. Multiplying these two equations on both sides and canceling the common factor $x_1x_2(\neq 0)$, we know $k_1k_2 = 4$. Since $k_1, k_2$ are positive integers, we have
$$(k_1, k_2) = (2, 2), (1, 4), \text{ or } (4, 1).$$
Therefore $S_2 = 4$ or $5$, which is of course less than $3 \cdot 2 = 6$.

Now consider the case $n = 3$. Without loss of generality, we may assume that $x_3$ is the largest among $x_1, x_2, x_3$. Then
$$k_3 = \frac{x_1 + x_2}{x_3} \le \frac{x_3 + x_3}{x_3} = 2.$$
So $k_3 = 1$ or $2$; when $k_3 = 2$, the three numbers must be equal, and $S_3 = 6 < 9 = 3 \cdot 3$. Thus when $x_1, x_2, x_3$ are not all equal, we have $k_3 = 1$. Writing out the equations, we get
$$k_1x_1 = x_2 + x_3, \quad k_2x_2 = x_1 + x_3, \quad x_3 = x_1 + x_2.$$
From the first and third equations, we know $(k_1 - 1)x_1 = 2x_2$. Similarly, from the second and third equations, we know $(k_2 - 1)x_2 = 2x_1$. This is exactly the same as the case $n = 2$, and we can obtain
$$(k_1, k_2, k_3) = (3, 3, 1), (2, 5, 1), \text{ or } (5, 2, 1).$$
Hence $S_3 = 7$ or $8$, both of which are less than $3 \cdot 3 = 9$. This concludes the case $n = 3$.

Now consider the general case $n \ge 4$. If among these $x$'s, the largest number appears consecutively 2 or more times, then all the numbers must be equal. This is because if the $x_j$'s are not all equal, then there will always be some $x_i$ at one end such that one of its neighbors is the largest number while the other neighbor is a positive integer strictly less than $x_j$, so the sum of the left and right neighbors cannot possibly be divisible by $x_j$ (since the sum lies strictly between $x_j$ and $2x_j$). So under this assumption, $S_n = 2n < 3n$.

In the case where the largest numbers among the $x$'s all appear in isolation, let $x_j$ be one of these largest numbers. As in the case $n = 3$, we can deduce that
$$k_j = 1, \quad \text{i.e.} \quad x_j = x_{j-1} + x_{j+1}. \qquad (1)$$
The equations obtained from the two neighbors on either side are
$$k_{j-1}x_{j-1} = x_{j-2} + x_j, \quad k_{j+1}x_{j+1} = x_j + x_{j+2}.$$

Combining with (1), we get
$$(k_{j-1} - 1)x_{j-1} = x_{j-2} + x_{j+1}, \text{ and } (k_{j+1} - 1)x_{j+1} = x_{j+2} + x_{j-1}.$$
(Note that from the maximality of $x_j$, we know that the two neighboring multiples $k_{j-1}, k_{j+1}$ are both greater than 1.) Observe: when $x_j$ is deleted, the two neighbors of $x_{j-1}$ become $x_{j-2}$ and $x_{j+1}$; similarly, the two neighbors of $x_{j+1}$ become $x_{j+2}$ and $x_{j-1}$. This argument tells us that when one of the largest numbers $x_j$ is removed, the remaining $n-1$ numbers still satisfy the conditions of the problem (i.e., the sum of the left and right neighbors is a multiple of itself), and among the multiples, besides the one 1 (namely $k_j$) being removed, the two neighboring multiples (namely $k_{j-1}$ and $k_{j+1}$) each also decrease by 1. Therefore we obtain $S_{n-1} = S_n - 3$. Since we have already proved the cases $n=1,2,3$, the result for general positive integers $n$ follows by mathematical induction. Q.E.D.