Determine all sequences $a_0 , a_1 , a_2 , \ldots$ of positive integers with $a_0 \ge 2015$ such that for all integers $n\ge 1$:
   (i) $a_{n+2}$ is divisible by $a_n$ ;
   (ii) $|s_{n+1} - (n + 1)a_n | = 1$, where $s_{n+1} = a_{n+1} - a_n + a_{n-1} - \cdots + (-1)^{n+1} a_0$ .

[i]Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand[/i]
1. **Define the sequence and initial conditions:**
   Let \( \{a_n\} \) be a sequence of positive integers with \( a_0 \geq 2015 \). We need to satisfy the following conditions for all integers \( n \geq 1 \):
   - \( a_{n+2} \) is divisible by \( a_n \).
   - \( |s_{n+1} - (n + 1)a_n | = 1 \), where \( s_{n+1} = a_{n+1} - a_n + a_{n-1} - \cdots + (-1)^{n+1} a_0 \).

2. **Simplify the expression for \( s_{n+1} \):**
   We start by noting that:
   \[
   s_{n+1} = a_{n+1} - a_n + a_{n-1} - \cdots + (-1)^{n+1} a_0
   \]
   Adding \( s_{n+1} \) and \( s_n \):
   \[
   s_{n+1} + s_n = \left(a_{n+1} + \sum_{i=0}^{n} (-1)^{n+1-i}a_i\right) + \left(\sum_{i=0}^{n}(-1)^{n-i}a_i\right) = a_{n+1}
   \]
   This simplifies to:
   \[
   s_{n+1} + s_n = a_{n+1}
   \]

3. **Establish the recurrence relation:**
   From the given conditions, we have:
   \[
   s_{n} = na_{n-1} \pm 1 \quad \text{and} \quad s_{n+1} = (n+1)a_n \pm 1
   \]
   Adding these, we get:
   \[
   a_{n+1} = (n+1)a_n + na_{n-1} + c_{n+1}
   \]
   where \( c_{n+1} \in \{-2, 0, 2\} \).

4. **Analyze the divisibility condition:**
   Using the recurrence relation, we can derive:
   \[
   a_n \mid (n+2)(n-1) a_{n-2} + (n+2)c_n - (n+1)c_{n+1} - c_{n+2}
   \]
   This implies that \( \frac{a_n}{a_{n-2}} \) must be an integer.

5. **Bound the sequence:**
   We need to show that \( a_n \geq 4n + 9 \) for \( n \geq 1 \). We use induction:
   - Base case: For \( n = 1 \), \( a_1 \geq 4 \cdot 1 + 9 = 13 \).
   - Inductive step: Assume \( a_n \geq 4n + 9 \). Then:
     \[
     a_{n+1} \geq (n+1)a_n - 2 \geq (n+1)(4n + 9) - 2 = 4(n+1) + 9
     \]

6. **Determine the general form of the sequence:**
   Using the recurrence relation and the bounds, we find:
   \[
   a_{n+2} = (n+4)(n+1)a_n
   \]
   This implies:
   \[
   a_n = \frac{(n+2)!}{n+1} c
   \]
   for some constant \( c \).

7. **Verify the initial conditions:**
   Given \( a_0 \geq 2015 \), we need to check the initial terms:
   \[
   a_1 = 3c, \quad a_2 = 8c
   \]
   Using the condition \( |s_2 - 2a_1| = 1 \), we get:
   \[
   3a_1 - a_2 = a_0 \pm 1 \Rightarrow 3(3c) - 8c = a_0 \pm 1 \Rightarrow a_0 = c \pm 1
   \]

8. **Conclusion:**
   The sequences that satisfy the given conditions are:
   \[
   \boxed{a_0 = c-1, c \geq 2016, a_n = \frac{(n+2)!}{n+1}c, \forall n \geq 1}
   \]
   and
   \[
   \boxed{a_0 = c+1, c \geq 2014, a_n = \frac{(n+2)!}{n+1}c, \forall n \geq 1}
   \]