Farmer James has three types of cows on his farm. A cow with zero legs is called a $\textit{ground beef}$, a cow with one leg is called a $\textit{steak}$, and a cow with two legs is called a $\textit{lean beef}$. Farmer James counts a total of $20$ cows and $18$ legs on his farm. How many more $\textit{ground beef}$s than $\textit{lean beef}$s does Farmer James have?

[i]Proposed by James Lin[/i]
1. Let \( x \) be the number of ground beefs (cows with 0 legs), \( y \) be the number of steaks (cows with 1 leg), and \( z \) be the number of lean beefs (cows with 2 legs).
2. We are given two equations based on the problem statement:
   \[
   x + y + z = 20 \quad \text{(total number of cows)}
   \]
   \[
   0x + 1y + 2z = 18 \quad \text{(total number of legs)}
   \]
   Simplifying the second equation, we get:
   \[
   y + 2z = 18
   \]
3. We need to find how many more ground beefs than lean beefs Farmer James has, i.e., we need to find \( x - z \).
4. From the first equation, solve for \( y \):
   \[
   y = 20 - x - z
   \]
5. Substitute \( y \) from the above equation into the second equation:
   \[
   (20 - x - z) + 2z = 18
   \]
   Simplify:
   \[
   20 - x + z = 18
   \]
   \[
   z - x = -2
   \]
   \[
   x - z = 2
   \]
6. Therefore, the number of ground beefs is 2 more than the number of lean beefs.

\(\blacksquare\)

The final answer is \( \boxed{ 2 } \)