For positive integers $n, m$ define $f(n,m)$ as follows. Write a list of $ 2001$ numbers $a_i$, where $a_1 = m$, and $a_{k+1}$ is the residue of $a_k^2$  $mod \, n$ (for $k = 1, 2,..., 2000$). Then put $f(n,m) = a_1-a_2 + a_3 -a_4 + a_5- ... + a_{2001}$. For which $n \ge 5$ can we find m such that $2 \le  m \le n/2$ and $f(m,n) > 0$?
1. **Define the function \( f(n, m) \):**
   - We start with \( a_1 = m \).
   - For \( k = 1, 2, \ldots, 2000 \), we define \( a_{k+1} \) as the residue of \( a_k^2 \mod n \).
   - The function \( f(n, m) \) is then given by:
     \[
     f(n, m) = a_1 - a_2 + a_3 - a_4 + \cdots + a_{2001}
     \]

2. **Analyze the function \( f(n, a_2) \):**
   - Notice that \( a_2 = m^2 \mod n \).
   - We can express \( f(n, a_2) \) as:
     \[
     f(n, a_2) = a_2 - a_3 + a_4 - a_5 + \cdots + a_{2002}
     \]
   - Since \( a_{2002} = a_1 \mod n \), we have:
     \[
     f(n, a_2) = a_2 - a_3 + a_4 - a_5 + \cdots + a_{2001} - a_{2002} + a_{2002}
     \]
   - Simplifying, we get:
     \[
     f(n, a_2) = -f(n, m) + a_1 + a_{2002}
     \]
   - Since \( a_{2002} = a_1 \mod n \), we have:
     \[
     f(n, a_2) = -f(n, m) + 2a_1
     \]

3. **Evaluate the inequality:**
   - Given \( -f(n, m) \leq 0 \), it follows that:
     \[
     f(n, a_2) > 0
     \]
   - This implies that if \( f(n, m) \leq 0 \), then \( f(n, a_2) > 0 \).

4. **Consider specific values of \( m \) and \( a_2 \):**
   - Assume \( m \) and \( a_2 \) are both \( \leq n/2 \).
   - We can choose a positive result between \( f(n, m) \) and \( f(n, a_2) \).

5. **Apply this to specific values:**
   - For \( m = 2 \) and \( a_2 = 4 \), we can cover all \( n \geq 8 \).

6. **Evaluate smaller values of \( n \):**
   - Evaluate \( f(5, 2) \):
     \[
     f(5, 2) = 2 - 4 + 1 - 1 + 1 - 1 + \cdots + 1 = -1
     \]
     - This leads to \( n = 5 \) not working (since \( 2 \) is the only permissible value for \( m \)).
   - Evaluate \( f(6, 3) \):
     \[
     f(6, 3) = 3 - 3 + 3 - 3 + 3 - 3 + \cdots + 3 = 3
     \]
     - So \( n = 6 \) works.
   - Evaluate \( f(7, 3) \):
     \[
     f(7, 3) = 3 - 2 + 4 - 2 + 4 - 2 + \cdots + 4 = 2003
     \]
     - So \( n = 7 \) also works.

Conclusion:
For \( n \geq 5 \), the values of \( n \) for which we can find \( m \) such that \( 2 \leq m \leq n/2 \) and \( f(n, m) > 0 \) are \( n = 6 \) and \( n = 7 \).

The final answer is \( \boxed{ n = 6 } \) and \( n = 7 \).