Given that \begin{align*}x &= 1 - \frac 12 + \frac13 - \frac 14 + \cdots + \frac1{2007},\\ y &= \frac{1}{1005} + \frac{1}{1006} + \frac{1}{1007} + \cdots + \frac 1{2007},\end{align*}
find the value of $k$ such that \[x = y + \frac 1k.\]
1. First, let's rewrite the given series for \( x \) and \( y \):
   \[
   x = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2007}
   \]
   \[
   y = \frac{1}{1005} + \frac{1}{1006} + \cdots + \frac{1}{2007}
   \]

2. Notice that \( x \) is an alternating harmonic series. We can express \( x \) as:
   \[
   x = \sum_{n=1}^{2007} \frac{(-1)^{n+1}}{n}
   \]

3. To find \( y \), we sum the harmonic series from \( \frac{1}{1005} \) to \( \frac{1}{2007} \):
   \[
   y = \sum_{n=1005}^{2007} \frac{1}{n}
   \]

4. We need to find \( k \) such that:
   \[
   x = y + \frac{1}{k}
   \]

5. Let's consider the partial sums of the harmonic series. The alternating harmonic series up to \( n \) terms can be approximated by:
   \[
   \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \approx \ln(2)
   \]
   However, since we are summing up to 2007 terms, we need to consider the specific terms.

6. We can approximate the sum \( x \) by breaking it into two parts:
   \[
   x \approx \left( \sum_{n=1}^{2007} \frac{1}{n} \right) - 2 \left( \sum_{n=1}^{1003} \frac{1}{2n} \right)
   \]

7. The sum \( y \) is straightforward:
   \[
   y = \sum_{n=1005}^{2007} \frac{1}{n}
   \]

8. To find \( k \), we need to equate:
   \[
   \sum_{n=1}^{2007} \frac{(-1)^{n+1}}{n} = \sum_{n=1005}^{2007} \frac{1}{n} + \frac{1}{k}
   \]

9. By comparing the sums, we can see that the missing term to balance the equation is:
   \[
   \frac{1}{k} = \sum_{n=1}^{1004} \frac{1}{n} - \sum_{n=1}^{1004} \frac{1}{2n}
   \]

10. Simplifying the above, we get:
    \[
    \frac{1}{k} = \sum_{n=1}^{1004} \frac{1}{2n}
    \]

11. Therefore, \( k \) is approximately:
    \[
    k = 2 \times 1004 = 2008
    \]

The final answer is \( \boxed{ k = 2008 } \).