Compute the maximum real value of $a$ for which there is an integer $b$ such that $\frac{ab^2}{a+2b} = 2019$. Compute the maximum possible value of $a$.
1. We start with the given equation:
   \[
   \frac{ab^2}{a+2b} = 2019
   \]
   Multiply both sides by \(a + 2b\) to clear the fraction:
   \[
   ab^2 = 2019(a + 2b)
   \]
2. Distribute \(2019\) on the right-hand side:
   \[
   ab^2 = 2019a + 4038b
   \]
3. Rearrange the equation to isolate \(a\):
   \[
   ab^2 - 2019a = 4038b
   \]
   Factor out \(a\) from the left-hand side:
   \[
   a(b^2 - 2019) = 4038b
   \]
   Solve for \(a\):
   \[
   a = \frac{4038b}{b^2 - 2019}
   \]
4. To maximize \(a\), we need to minimize the denominator \(b^2 - 2019\). Since \(b\) is an integer, the smallest value \(b^2 - 2019\) can take is 1 (as \(b^2\) must be greater than 2019 for \(a\) to be positive). Therefore, we set:
   \[
   b^2 - 2019 = 1 \implies b^2 = 2020
   \]
   However, 2020 is not a perfect square, so we need to find the closest integer values for \(b\) that minimize \(b^2 - 2019\).

5. The closest perfect squares to 2019 are \(44^2 = 1936\) and \(45^2 = 2025\). We choose \(b = 45\) because it is closer to 2019 and will give a smaller denominator:
   \[
   b = 45 \implies b^2 = 2025
   \]
   Substitute \(b = 45\) into the equation for \(a\):
   \[
   a = \frac{4038 \cdot 45}{45^2 - 2019} = \frac{4038 \cdot 45}{2025 - 2019} = \frac{4038 \cdot 45}{6}
   \]
6. Simplify the expression:
   \[
   a = \frac{4038 \cdot 45}{6} = 673 \cdot 45 = 30285
   \]

The final answer is \(\boxed{30285}\).