Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$.  For how many values of $x$ is $f_{100}(x)=0$?

$\textbf{(A) }299\qquad
\textbf{(B) }300\qquad
\textbf{(C) }301\qquad
\textbf{(D) }302\qquad
\textbf{(E) }303\qquad$
1. **Define the function \( f_0(x) \):**
   \[
   f_0(x) = x + |x - 100| - |x + 100|
   \]
   We need to analyze this function in different intervals of \( x \).

2. **Analyze \( f_0(x) \) in different intervals:**
   - For \( x < -100 \):
     \[
     f_0(x) = x + (100 - x) - (-100 - x) = x + 100 - x + 100 + x = x + 200
     \]
   - For \( -100 \leq x \leq 100 \):
     \[
     f_0(x) = x + (100 - x) - (x + 100) = x + 100 - x - x - 100 = -x
     \]
   - For \( x > 100 \):
     \[
     f_0(x) = x + (x - 100) - (x + 100) = x + x - 100 - x - 100 = x - 200
     \]

   Therefore, we have:
   \[
   f_0(x) = \begin{cases}
   x + 200 & \text{if } x < -100 \\
   -x & \text{if } -100 \leq x \leq 100 \\
   x - 200 & \text{if } x > 100
   \end{cases}
   \]

3. **Define the recursive function \( f_n(x) \):**
   \[
   f_n(x) = |f_{n-1}(x)| - 1
   \]

4. **Determine the possible values of \( f_{100}(x) \):**
   We need \( f_{100}(x) = 0 \). This implies that \( f_{99}(x) = 1 \), \( f_{98}(x) = 2 \), and so on, up to \( f_0(x) \).

5. **Identify the range of \( f_0(x) \):**
   - For \( x < -100 \), \( f_0(x) = x + 200 \). The values range from \( -300 \) to \( -102 \) (inclusive).
   - For \( -100 \leq x \leq 100 \), \( f_0(x) = -x \). The values range from \( -100 \) to \( 100 \) (inclusive).
   - For \( x > 100 \), \( f_0(x) = x - 200 \). The values range from \( -98 \) to \( 300 \) (inclusive).

   Combining these ranges, \( f_0(x) \) can take any value from \( -300 \) to \( 300 \) in steps of 2.

6. **Count the number of values \( x \) such that \( f_0(x) \) is in the range \( -300 \) to \( 300 \):**
   The sequence \( -300, -298, \ldots, 300 \) is an arithmetic sequence with a common difference of 2. The number of terms in this sequence is:
   \[
   \frac{300 - (-300)}{2} + 1 = \frac{600}{2} + 1 = 301
   \]

Conclusion:
\[
\boxed{301}
\]