Jacob and Laban take turns playing a game. Each of them starts with the list of square numbers $1, 4, 9, \dots, 2021^2$, and there is a whiteboard in front of them with the number $0$ on it. Jacob chooses a number $x^2$ from his list, removes it from his list, and replaces the number $W$ on the whiteboard with $W + x^2$. Laban then does the same with a number from his list, and the repeat back and forth until both of them have no more numbers in their list. Now every time that the number on the whiteboard is divisible by $4$ after a player has taken his turn, Jacob gets a sheep. Jacob wants to have as many sheep as possible. What is the greatest number $K$ such that Jacob can guarantee to get at least $K$ sheep by the end of the game, no matter how Laban plays?
1. **Understanding the Problem:**
   - Jacob and Laban take turns choosing square numbers from their respective lists and adding them to a number on the whiteboard.
   - Jacob gets a sheep every time the number on the whiteboard is divisible by 4 after his turn.
   - We need to determine the maximum number of sheep Jacob can guarantee to get, regardless of Laban's strategy.

2. **Analyzing the Square Numbers Modulo 4:**
   - The square numbers modulo 4 are either 0 or 1.
   - Specifically, \( x^2 \equiv 0 \pmod{4} \) if \( x \) is even, and \( x^2 \equiv 1 \pmod{4} \) if \( x \) is odd.

3. **Initial Setup:**
   - Both players start with the list of square numbers \( 1^2, 2^2, 3^2, \ldots, 2021^2 \).
   - There are 1011 odd squares (which are 1 modulo 4) and 1010 even squares (which are 0 modulo 4).

4. **Total Sum of Squares:**
   - The total sum of all the squares is \( 1^2 + 2^2 + 3^2 + \ldots + 2021^2 \).
   - Modulo 4, this sum is \( 1011 \times 1 + 1010 \times 0 = 1011 \).

5. **Guaranteed Sheep Calculation:**
   - The total sum modulo 4 is 1011, which is 3 modulo 4.
   - The game will pass through multiples of 4 as the sum increases.
   - Since the total sum is 2022, the game will pass through the numbers \( 4, 8, 12, \ldots, 2020 \), which are 505 multiples of 4.
   - Jacob can guarantee at least 505 sheep from these multiples.

6. **Jacob's Strategy:**
   - Jacob can start by playing a 0 (an even square), ensuring the initial number on the whiteboard is 0.
   - This guarantees an additional sheep at the start, making the total at least 506 sheep.

7. **Laban's Counter-Strategy:**
   - If Jacob starts with 0, Laban can respond with 1.
   - Laban can then mirror Jacob's moves to ensure that the number on the whiteboard is never a multiple of 4 after Laban's turn.
   - This strategy ensures that Jacob only gets sheep when the total sum is a multiple of 4, which happens 506 times.

8. **Conclusion:**
   - Jacob can guarantee at least 506 sheep by starting with a 0 and ensuring the game passes through multiples of 4.

The final answer is \(\boxed{506}\).