1. **Substitute $m$ with $n - k$ in the inequalities $k \leq 2m$ and $m \leq 2k$:**

Given $k + m = n$, we can replace $m$ with $n - k$. Thus, the inequalities become:
$$k \leq 2(n - k) \quad \text{and} \quad n - k \leq 2k$$
Simplifying these inequalities:
$$k \leq 2n - 2k \implies 3k \leq 2n \implies k \leq \frac{2n}{3}$$
$$n - k \leq 2k \implies n \leq 3k \implies k \geq \frac{n}{3}$$

2. **Determine the range of $k$:**

Since $k$ is an integer, it can range from $\lceil \frac{n}{3} \rceil$ to $\lfloor \frac{2n}{3} \rfloor$, inclusive.

3. **Analyze the sum $x_nx_1x_2 + x_1x_2x_3 + \cdots + x_{n-1}x_nx_1$:**

Each term in the sum is either $a^2b$ or $ab^2$ due to condition (iii) that no three consecutive terms are equal. Let $y$ be the number of terms $a^2b$ and $n - y$ be the number of terms $ab^2$. Therefore, the sum can be expressed as:
$$(a^2b)y + (ab^2)(n - y)$$

4. **Count the number of $a$'s in the sum:**

Each $a$ is counted three times in the sum. Since there are $k$ terms equal to $a$, the total number of $a$'s in the sum is $3k$. Thus, we have:
$$2y + (n - y) = 3k \implies y = 3k - n$$

5. Determine the possible values for the sum:

Since $y = 3k - n$ and $k$ ranges from $\lceil \frac{n}{3} \rceil$ to $\lfloor \frac{2n}{3} \rfloor$, the number of possible values for $y$ is equal to the number of possible values for $k$. Therefore, the number of possible values for the sum is:
$$\lfloor \frac{2n}{3} \rfloor - \lceil \frac{n}{3} \rceil + 1$$