The numbers $3$, $4$ ,$...$, $2019$ are written on a blackboard. David and Edgardo play alternately, starting with David. On their turn, each player must erase a number from the board and write two positive integers whose sum is
equal to the number just deleted. The winner is the one who makes all the numbers on the board equal. Determine who has a winning strategy and describe it.
1. **Initial Setup and Claim**:
   - The numbers $3, 4, \ldots, 2019$ are written on a blackboard.
   - David and Edgardo play alternately, starting with David.
   - On each turn, a player erases a number $n$ and writes two positive integers $a$ and $b$ such that $a + b = n$.
   - The game ends when all numbers on the board are equal.
   - Claim: Edgardo always wins.

2. **Reduction of Numbers**:
   - When a number $n$ is removed and replaced by $a$ and $b$ such that $a + b = n$ and $a, b > 0$, it follows that $a, b < n$.
   - This implies that the smallest number on the board will always be less than or equal to $3$.

3. **Possible Final Situations**:
   - The game ends when all numbers on the board are equal.
   - Possible final situations:
     - All numbers are $3$.
     - All numbers are $2$.
     - All numbers are $1$.

4. **Analysis of Final Situations**:
   - **Situation 1: All numbers are $3$**:
     - To achieve this, the number $4$ must be removed and replaced by $(2, 2)$ or $(1, 3)$.
     - Both replacements introduce a number smaller than $3$, so this situation is impossible.
   - **Situation 2: All numbers are $2$**:
     - To achieve this, the number $3$ must be removed and replaced by $(1, 2)$.
     - This introduces a number smaller than $2$, so this situation is impossible.
   - **Situation 3: All numbers are $1$**:
     - This is the only situation that does not lead to a contradiction.

5. **Turns Required to Reach Final Situation**:
   - Number $3$ requires $2$ turns to be replaced by three $1$'s.
   - Number $4$ requires $3$ turns to be replaced by four $1$'s.
   - By induction, any number $m$ requires $m-1$ turns to be replaced by $m$ $1$'s.
   - Total turns required:
     \[
     (3-1) + (4-1) + (5-1) + \ldots + (2019-1)
     \]
   - Simplifying the sum:
     \[
     \sum_{n=3}^{2019} (n-1) = \sum_{n=3}^{2019} n - \sum_{n=3}^{2019} 1 = \left( \sum_{n=3}^{2019} n \right) - 2017
     \]
   - The sum of the first $2019$ natural numbers is:
     \[
     \sum_{n=1}^{2019} n = \frac{2019 \cdot 2020}{2} = 2039190
     \]
   - The sum from $3$ to $2019$ is:
     \[
     \sum_{n=3}^{2019} n = 2039190 - 3 = 2039187
     \]
   - Subtracting $2017$:
     \[
     2039187 - 2017 = 2037170
     \]
   - Since $2037170$ is even, the last turn will be Edgardo's.

Conclusion:
\[
\boxed{\text{Edgardo always wins.}}
\]