For some positive integer $n$, there exists $n$ different positive integers $a_1, a_2, ..., a_n$ such that
$(1)$ $a_1=1, a_n=2000$
$(2)$ $\forall i\in \mathbb{Z}$  $s.t.$  $2\le i\le n, a_i -a_{i-1}\in \{-3,5\}$
Determine the maximum value of n.
1. We start with the given conditions:
   - \( a_1 = 1 \)
   - \( a_n = 2000 \)
   - For all \( i \) such that \( 2 \le i \le n \), \( a_i - a_{i-1} \in \{-3, 5\} \)

2. We need to determine the maximum value of \( n \). First, let's analyze the possible values of \( a_i \):
   - If \( a_i - a_{i-1} = 5 \), then \( a_i \) increases by 5.
   - If \( a_i - a_{i-1} = -3 \), then \( a_i \) decreases by 3.

3. We observe that the sequence can be written as:
   \[
   a_i = a_1 + \sum_{j=2}^{i} (a_j - a_{j-1})
   \]
   Since \( a_1 = 1 \), we have:
   \[
   a_i = 1 + \sum_{j=2}^{i} (a_j - a_{j-1})
   \]

4. Let \( x \) be the number of times we add 5 and \( y \) be the number of times we subtract 3. Then:
   \[
   x + y = n - 1
   \]
   and
   \[
   1 + 5x - 3y = 2000
   \]

5. Solving the equation \( 1 + 5x - 3y = 2000 \):
   \[
   5x - 3y = 1999
   \]

6. We need to find non-negative integer solutions for \( x \) and \( y \). Rearranging the equation:
   \[
   5x = 1999 + 3y
   \]
   \[
   x = \frac{1999 + 3y}{5}
   \]

7. For \( x \) to be an integer, \( 1999 + 3y \) must be divisible by 5. We check the congruence:
   \[
   1999 + 3y \equiv 0 \pmod{5}
   \]
   \[
   1999 \equiv -1 \pmod{5}
   \]
   \[
   -1 + 3y \equiv 0 \pmod{5}
   \]
   \[
   3y \equiv 1 \pmod{5}
   \]
   \[
   y \equiv 2 \pmod{5}
   \]

8. Let \( y = 5k + 2 \) for some integer \( k \). Substituting \( y \) back into the equation for \( x \):
   \[
   x = \frac{1999 + 3(5k + 2)}{5}
   \]
   \[
   x = \frac{1999 + 15k + 6}{5}
   \]
   \[
   x = \frac{2005 + 15k}{5}
   \]
   \[
   x = 401 + 3k
   \]

9. Therefore, the total number of terms \( n \) is:
   \[
   n = x + y + 1 = (401 + 3k) + (5k + 2) + 1
   \]
   \[
   n = 404 + 8k
   \]

10. To maximize \( n \), we need to find the largest \( k \) such that \( n \leq 2000 \):
    \[
    404 + 8k \leq 2000
    \]
    \[
    8k \leq 1596
    \]
    \[
    k \leq 199.5
    \]
    Since \( k \) must be an integer, the largest possible value for \( k \) is 199.

11. Substituting \( k = 199 \) back into the equation for \( n \):
    \[
    n = 404 + 8 \cdot 199
    \]
    \[
    n = 404 + 1592
    \]
    \[
    n = 1996
    \]

Therefore, the maximum value of \( n \) is \( 1996 \).

The final answer is \( \boxed{1996} \).