Determine all positive integers that cannot be written as $\frac{a}{b} + \frac{a+1}{b+1}$ where $a$ and $b$ are positive integers.
1. We start with the given expression:
   \[
   \frac{a}{b} + \frac{a+1}{b+1}
   \]
   We need to determine all positive integers that cannot be written in this form where \(a\) and \(b\) are positive integers.

2. Combine the fractions:
   \[
   \frac{a}{b} + \frac{a+1}{b+1} = \frac{a(b+1) + (a+1)b}{b(b+1)} = \frac{ab + a + ab + b}{b(b+1)} = \frac{2ab + a + b}{b(b+1)}
   \]

3. Simplify the expression:
   \[
   \frac{2ab + a + b}{b(b+1)} = \frac{2ab + a + b}{b^2 + b}
   \]

4. Suppose a positive integer \(n\) can be written as:
   \[
   n = \frac{2ab + a + b}{b^2 + b}
   \]

5. Rearrange the equation:
   \[
   n(b^2 + b) = 2ab + a + b
   \]

6. Simplify further:
   \[
   nb^2 + nb = 2ab + a + b
   \]

7. Rearrange to isolate terms involving \(a\):
   \[
   nb^2 + nb - b = 2ab + a
   \]

8. Factor out \(a\) on the right-hand side:
   \[
   nb^2 + b(n - 1) = a(2b + 1)
   \]

9. Solve for \(a\):
   \[
   a = \frac{nb^2 + b(n - 1)}{2b + 1}
   \]

10. For \(a\) to be an integer, the numerator \(nb^2 + b(n - 1)\) must be divisible by \(2b + 1\). This implies:
    \[
    2b + 1 \mid nb^2 + b(n - 1)
    \]

11. Consider the divisibility condition:
    \[
    2b + 1 \mid n(b^2 + b) - b
    \]

12. Simplify the condition:
    \[
    2b + 1 \mid n(b^2 + b) - b = n(b^2 + b) - b = nb^2 + nb - b
    \]

13. Since \(2b + 1\) is an odd number, \(n - 2\) must be divisible by an odd number greater than 1. Therefore, \(n\) must be of the form:
    \[
    n = 2 + k(2b + 1)
    \]
    for some integer \(k \geq 0\).

14. The set of all positive integers that cannot be written as \(\frac{a}{b} + \frac{a+1}{b+1}\) is:
    \[
    S = \{1\} \cup \{2^m + 2 : m \in \mathbb{N}_0 \}
    \]

The final answer is \( \boxed{ \{1\} \cup \{2^m + 2 : m \in \mathbb{N}_0 \} } \).