Alice and Bob play a game on a Cartesian Coordinate Plane. At the beginning, Alice chooses a lattice point $ \left(x_{0}, y_{0}\right) $ and places a pudding. Then they plays by turns (B goes first) according to the rules

a. If $ A $ places a pudding on $ \left(x,y\right) $ in the last round, then $ B $ can only place a pudding on one of $ \left(x+2, y+1\right), \left(x+2, y-1\right), \left(x-2, y+1\right), \left(x-2, y-1\right) $ 

b. If $ B $ places a pudding on $ \left(x,y\right) $ in the last round, then $ A $ can only place a pudding on one of $ \left(x+1, y+2\right), \left(x+1, y-2\right), \left(x-1, y+2\right), \left(x-1, y-2\right) $

Furthermore, if there is already a pudding on $ \left(a,b\right) $, then no one can place a pudding on $ \left(c,d\right) $ where $ c \equiv a \pmod{n}, d \equiv b \pmod{n} $.

1. Who has a winning strategy when $ n = 2018 $

1. Who has a winning strategy when $ n = 2019 $
To determine who has a winning strategy, we need to analyze the movement rules and the constraints given by the modulo condition. Let's break down the problem step by step.

### Case 1: \( n = 2018 \)

1. **Initial Setup and Movement Analysis:**
   - Alice places a pudding at \((x_0, y_0)\).
   - Bob can move to \((x_0 + 2, y_0 + 1)\), \((x_0 + 2, y_0 - 1)\), \((x_0 - 2, y_0 + 1)\), or \((x_0 - 2, y_0 - 1)\).
   - Alice can then move to \((x_1 + 1, y_1 + 2)\), \((x_1 + 1, y_1 - 2)\), \((x_1 - 1, y_1 + 2)\), or \((x_1 - 1, y_1 - 2)\).

2. **Checkerboard Coloring and Correspondence:**
   - Color the grid like a checkerboard. Each move by Bob changes the parity of both coordinates (since he moves by \((\pm 2, \pm 1)\)), and each move by Alice also changes the parity of both coordinates (since she moves by \((\pm 1, \pm 2)\)).
   - This means that if Alice starts on a white square, Bob will always move to a black square, and vice versa.

3. **Modulo Condition:**
   - The modulo condition \(c \equiv a \pmod{2018}\) and \(d \equiv b \pmod{2018}\) implies that if a square \((a, b)\) is occupied, then all squares \((a + k \cdot 2018, b + l \cdot 2018)\) for integers \(k\) and \(l\) are also considered occupied.
   - Given the large size of \(n = 2018\), the grid effectively wraps around every 2018 units in both directions.

4. **Bob's Strategy:**
   - Bob can always move to a corresponding black square that is guaranteed to be empty because the grid is large enough (2018 units) to ensure that the corresponding black square is not already occupied by the modulo condition.
   - Therefore, Bob can always find a valid move, ensuring that he can continue to place puddings without running out of space.

### Case 2: \( n = 2019 \)

1. **Initial Setup and Movement Analysis:**
   - Similar to the previous case, Alice places a pudding at \((x_0, y_0)\).
   - Bob can move to \((x_0 + 2, y_0 + 1)\), \((x_0 + 2, y_0 - 1)\), \((x_0 - 2, y_0 + 1)\), or \((x_0 - 2, y_0 - 1)\).
   - Alice can then move to \((x_1 + 1, y_1 + 2)\), \((x_1 + 1, y_1 - 2)\), \((x_1 - 1, y_1 + 2)\), or \((x_1 - 1, y_1 - 2)\).

2. **Checkerboard Coloring and Correspondence:**
   - Color the grid like a checkerboard. Each move by Bob changes the parity of both coordinates, and each move by Alice also changes the parity of both coordinates.
   - This means that if Alice starts on a white square, Bob will always move to a black square, and vice versa.

3. **Modulo Condition:**
   - The modulo condition \(c \equiv a \pmod{2019}\) and \(d \equiv b \pmod{2019}\) implies that if a square \((a, b)\) is occupied, then all squares \((a + k \cdot 2019, b + l \cdot 2019)\) for integers \(k\) and \(l\) are also considered occupied.
   - Given the large size of \(n = 2019\), the grid effectively wraps around every 2019 units in both directions.

4. **Bob's Strategy:**
   - Bob can always move to a corresponding black square that is guaranteed to be empty because the grid is large enough (2019 units) to ensure that the corresponding black square is not already occupied by the modulo condition.
   - Therefore, Bob can always find a valid move, ensuring that he can continue to place puddings without running out of space.

Conclusion:
- For both \(n = 2018\) and \(n = 2019\), Bob has a winning strategy by always moving to a corresponding black square that is guaranteed to be empty.

The final answer is Bob has a winning strategy for both \( \boxed{ n = 2018 } \) and \(n = 2019\).