The fraction $\frac1{10}$ can be expressed as the sum of two unit fraction in many ways, for example, $\frac1{30}+\frac1{15}$ and $\frac1{60}+\frac1{12}$.

Find the number of ways that $\frac1{2007}$ can be expressed as the sum of two distinct positive unit fractions.
1. We start with the given equation:
   \[
   \frac{1}{2007} = \frac{1}{m} + \frac{1}{n}
   \]
   where \(m\) and \(n\) are distinct positive integers.

2. Rewrite the equation with a common denominator:
   \[
   \frac{1}{2007} = \frac{m + n}{mn}
   \]

3. Cross-multiplying to clear the fractions, we get:
   \[
   mn = 2007(m + n)
   \]

4. Rearrange the equation to:
   \[
   mn - 2007m - 2007n = 0
   \]

5. Add \(2007^2\) to both sides to factorize:
   \[
   mn - 2007m - 2007n + 2007^2 = 2007^2
   \]

6. This can be rewritten as:
   \[
   (m - 2007)(n - 2007) = 2007^2
   \]

7. The number of ways to express \(\frac{1}{2007}\) as the sum of two distinct positive unit fractions corresponds to finding the number of distinct pairs \((m, n)\) such that \((m - 2007)(n - 2007) = 2007^2\).

8. The prime factorization of \(2007\) is:
   \[
   2007 = 3^2 \cdot 223
   \]
   Therefore:
   \[
   2007^2 = (3^2 \cdot 223)^2 = 3^4 \cdot 223^2
   \]

9. The number of divisors of \(2007^2\) is given by:
   \[
   (4 + 1)(2 + 1) = 5 \cdot 3 = 15
   \]

10. Each divisor \(d\) of \(2007^2\) gives a pair \((m, n)\) where:
    \[
    m - 2007 = d \quad \text{and} \quad n - 2007 = \frac{2007^2}{d}
    \]
    or vice versa.

11. Since \(m\) and \(n\) must be distinct, we exclude the case where \(m = n\). This leaves us with \(14\) distinct pairs.

12. Since the order of \(m\) and \(n\) does not matter, we divide the number of pairs by \(2\):
    \[
    \frac{14}{2} = 7
    \]

Conclusion:
\[
\boxed{7}
\]