If $a$ and $b$ are positive integers such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$, what is the greatest possible value of $a+b$?
1. Start with the given equation:
   \[
   \frac{1}{a} + \frac{1}{b} = \frac{1}{9}
   \]

2. Combine the fractions on the left-hand side:
   \[
   \frac{a + b}{ab} = \frac{1}{9}
   \]

3. Cross-multiply to eliminate the fractions:
   \[
   9(a + b) = ab
   \]

4. Rearrange the equation to set it to zero:
   \[
   ab - 9a - 9b = 0
   \]

5. Add 81 to both sides to factorize using Simon's Favorite Factoring Trick (SFFT):
   \[
   ab - 9a - 9b + 81 = 81
   \]

6. Factor the left-hand side:
   \[
   (a - 9)(b - 9) = 81
   \]

7. Identify the factor pairs of 81 to find possible values for \(a\) and \(b\):
   \[
   81 = 1 \times 81, \quad 3 \times 27, \quad 9 \times 9
   \]

8. Calculate \(a\) and \(b\) for each factor pair:
   - For \((a-9) = 1\) and \((b-9) = 81\):
     \[
     a = 1 + 9 = 10, \quad b = 81 + 9 = 90
     \]
     \[
     a + b = 10 + 90 = 100
     \]

   - For \((a-9) = 3\) and \((b-9) = 27\):
     \[
     a = 3 + 9 = 12, \quad b = 27 + 9 = 36
     \]
     \[
     a + b = 12 + 36 = 48
     \]

   - For \((a-9) = 9\) and \((b-9) = 9\):
     \[
     a = 9 + 9 = 18, \quad b = 9 + 9 = 18
     \]
     \[
     a + b = 18 + 18 = 36
     \]

9. Determine the greatest possible value of \(a + b\):
   \[
   \max(100, 48, 36) = 100
   \]

The final answer is \(\boxed{100}\).