If $x_1=5, x_2=401$, and 
\[
    x_n=x_{n-2}-\frac 1{x_{n-1}}
\]
for every $3\leq n \leq m$, what is the largest value of $m$?

$ 
\textbf{(A)}\ 406
\qquad\textbf{(B)}\ 2005
\qquad\textbf{(C)}\ 2006
\qquad\textbf{(D)}\ 2007
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Given the recurrence relation:
   \[
   x_n = x_{n-2} - \frac{1}{x_{n-1}}
   \]
   for \(3 \leq n \leq m\), with initial conditions \(x_1 = 5\) and \(x_2 = 401\).

2. We start by manipulating the recurrence relation. Multiply both sides by \(x_{n-1}\):
   \[
   x_n x_{n-1} = x_{n-2} x_{n-1} - 1
   \]

3. Rearrange the equation:
   \[
   x_n x_{n-1} - x_{n-1} x_{n-2} = -1
   \]

4. Notice that this equation can be summed over \(n\). Consider the sum from \(n = 3\) to \(n = m\):
   \[
   \sum_{n=3}^m (x_n x_{n-1} - x_{n-1} x_{n-2}) = \sum_{n=3}^m -1
   \]

5. The left-hand side is a telescoping series:
   \[
   (x_3 x_2 - x_2 x_1) + (x_4 x_3 - x_3 x_2) + \cdots + (x_m x_{m-1} - x_{m-1} x_{m-2})
   \]
   Most terms cancel out, leaving:
   \[
   x_m x_{m-1} - x_2 x_1
   \]

6. The right-hand side simplifies to:
   \[
   \sum_{n=3}^m -1 = -(m-2)
   \]

7. Equate the two sides:
   \[
   x_m x_{m-1} - x_2 x_1 = -(m-2)
   \]

8. Substitute the initial conditions \(x_1 = 5\) and \(x_2 = 401\):
   \[
   x_m x_{m-1} - 401 \cdot 5 = -(m-2)
   \]
   \[
   x_m x_{m-1} - 2005 = -(m-2)
   \]
   \[
   x_m x_{m-1} = 2007 - m
   \]

9. To find the largest value of \(m\), note that \(x_m x_{m-1}\) must be a valid product of real numbers. The sequence will break when \(x_m x_{m-1} = 0\), which happens when:
   \[
   2007 - m = 0
   \]
   \[
   m = 2007
   \]

10. For \(m = 2007\), we have \(x_{2007} x_{2006} = 0\). This implies \(x_{2007} = 0\) or \(x_{2006} = 0\). If \(x_{2007} = 0\), then \(x_{2008}\) is undefined because it involves division by zero.

Conclusion:
The largest value of \(m\) is \(2007\).

The final answer is \(\boxed{2007}\)