How many pairs $ (m,n)$ of positive integers with $ m < n$ fulfill the equation $ \frac {3}{2008} \equal{} \frac 1m \plus{} \frac 1n$?
1. Start with the given equation:
   \[
   \frac{3}{2008} = \frac{1}{m} + \frac{1}{n}
   \]
   To combine the fractions on the right-hand side, find a common denominator:
   \[
   \frac{3}{2008} = \frac{m + n}{mn}
   \]
   Cross-multiplying gives:
   \[
   3mn = 2008(m + n)
   \]

2. Rearrange the equation to isolate terms involving \(m\) and \(n\):
   \[
   3mn - 2008m - 2008n = 0
   \]
   Add \(2008^2\) to both sides:
   \[
   3mn - 2008m - 2008n + 2008^2 = 2008^2
   \]
   Factor the left-hand side:
   \[
   (3m - 2008)(3n - 2008) = 2008^2
   \]

3. Let \(a = 3m - 2008\) and \(b = 3n - 2008\). Then:
   \[
   ab = 2008^2
   \]
   Since \(m\) and \(n\) are positive integers and \(m < n\), we have \(a < b\).

4. We need \(a\) and \(b\) to be such that \(3m - 2008\) and \(3n - 2008\) are integers. This implies:
   \[
   3m \equiv 2008 \pmod{3} \quad \text{and} \quad 3n \equiv 2008 \pmod{3}
   \]
   Since \(2008 \equiv 2 \pmod{3}\), we have:
   \[
   3m \equiv 2 \pmod{3} \quad \text{and} \quad 3n \equiv 2 \pmod{3}
   \]
   Therefore:
   \[
   m \equiv 2 \pmod{3} \quad \text{and} \quad n \equiv 2 \pmod{3}
   \]

5. We need to find pairs \((a, b)\) such that \(a \equiv -1 \pmod{3}\) and \(b \equiv -1 \pmod{3}\). Since \(a = 3m - 2008\) and \(b = 3n - 2008\), we have:
   \[
   a \equiv -1 \pmod{3} \quad \text{and} \quad b \equiv -1 \pmod{3}
   \]

6. The number \(2008\) can be factored as:
   \[
   2008 = 2^3 \times 251
   \]
   Therefore, the number of divisors of \(2008^2\) is:
   \[
   (3+1)(1+1) = 8 \times 2 = 16
   \]
   We need to find the divisors of \(2008^2\) that are congruent to \(-1 \pmod{3}\). 

7. The divisors of \(2008^2\) that are congruent to \(-1 \pmod{3}\) are:
   \[
   1, 2, 4, 8, 16, 32, 64, 128, 251, 502, 1004, 2008, 4016, 8032, 16064, 32128
   \]
   Out of these, the ones that are \(\equiv -1 \pmod{3}\) are:
   \[
   2, 8, 32, 128, 502, 2008, 8032, 32128
   \]
   There are 8 such numbers.

8. Since \(a < b\), we need to count the pairs \((a, b)\) such that \(a < b\). There are 4 such pairs.

Conclusion:
\[
\boxed{4}
\]