Determine all functions $ f$ mapping the set of positive integers to the set of non-negative integers satisfying the following conditions:
(1) $ f(mn) \equal{} f(m)\plus{}f(n)$,
(2) $ f(2008) \equal{} 0$, and
(3) $ f(n) \equal{} 0$ for all $ n \equiv 39\pmod {2008}$.
1. **Given Conditions and Initial Observations:**
   - We are given a function \( f \) mapping the set of positive integers to the set of non-negative integers.
   - The function satisfies:
     1. \( f(mn) = f(m) + f(n) \) for all positive integers \( m \) and \( n \).
     2. \( f(2008) = 0 \).
     3. \( f(n) = 0 \) for all \( n \equiv 39 \pmod{2008} \).

2. **Analyzing Condition (2):**
   - From \( f(2008) = 0 \), we know that \( f \) evaluated at 2008 is zero.
   - Since \( f \) is non-negative, \( f(2008) = 0 \) implies that \( f \) must be zero for any divisor of 2008. This is because if \( 2008 = mn \), then \( 0 = f(2008) = f(m) + f(n) \), and since \( f(m) \) and \( f(n) \) are non-negative, both must be zero.

3. **Divisors of 2008:**
   - The prime factorization of 2008 is \( 2008 = 2^3 \times 251 \).
   - The divisors of 2008 are \( 1, 2, 4, 8, 251, 502, 1004, 2008 \).
   - Therefore, \( f(d) = 0 \) for each divisor \( d \) of 2008.

4. **Analyzing Condition (3):**
   - For \( n \equiv 39 \pmod{2008} \), we have \( n = 2008k + 39 \) for some integer \( k \).
   - Given \( f(n) = 0 \) for all such \( n \), we need to show that this implies \( f \) is zero for all positive integers.

5. **Coprime Argument:**
   - Suppose \( m \) is any positive integer coprime with 2008. Then there exists an integer \( a \) such that \( am \equiv 1 \pmod{2008} \) (by Bezout's identity).
   - Consider \( n = 39a \). Then \( nm \equiv 39 \pmod{2008} \), so \( f(nm) = f(39a \cdot m) = f(39) + f(a) + f(m) \).
   - Since \( nm \equiv 39 \pmod{2008} \), we have \( f(nm) = 0 \). Thus, \( 0 = f(39) + f(a) + f(m) \).
   - Given \( f(39) = 0 \) (from condition 3), we have \( 0 = f(a) + f(m) \). Since \( f(a) \geq 0 \) and \( f(m) \geq 0 \), it follows that \( f(a) = 0 \) and \( f(m) = 0 \).

6. **General Case:**
   - If \( m \) is not coprime with 2008, let \( d = \gcd(m, 2008) \). Then \( m = da \) where \( \gcd(a, 2008) = 1 \).
   - We have \( f(m) = f(da) = f(d) + f(a) \).
   - Since \( d \) is a divisor of 2008, \( f(d) = 0 \). Also, since \( a \) is coprime with 2008, \( f(a) = 0 \).
   - Therefore, \( f(m) = 0 + 0 = 0 \).

7. **Conclusion:**
   - We have shown that \( f(n) = 0 \) for all positive integers \( n \).

The final answer is \( \boxed{ f(n) = 0 } \) for all positive integers \( n \).