Solution:
Let $z_{1}, \ldots, z_{5}$ be the roots of $Q(z) = z^{5} + 2004 z - 1$. We can check these are distinct (by using the fact that there's one in a small neighborhood of each root of $z^{5} + 2004 z$, or by noting that $Q(z)$ is relatively prime to its derivative). And certainly none of the roots of $Q$ is the negative of another, since $z^{5} + 2004 z = 1$ implies $(-z)^{5} + 2004(-z) = -1$, so their squares are distinct as well. Then, $z_{1}^{2}, \ldots, z_{5}^{2}$ are the roots of $P$, so if we write $C$ for the leading coefficient of $P$, we have
$$\begin{aligned} \frac{P(1)}{P(-1)} & = \frac{C\left(1-z_{1}^{2}\right) \cdots\left(1-z_{5}^{2}\right)}{C\left(-1-z_{1}^{2}\right) \cdots\left(-1-z_{5}^{2}\right)} \\ & = \frac{\left[\left(1-z_{1}\right) \cdots\left(1-z_{5}\right)\right] \cdot \left[\left(1+z_{1}\right) \cdots\left(1+z_{5}\right)\right]}{\left[\left(i-z_{1}\right) \cdots\left(i-z_{5}\right)\right] \cdot \left[\left(i+z_{1}\right) \cdots\left(i+z_{5}\right)\right]} \\ & = \frac{\left[\left(1-z_{1}\right) \cdots\left(1-z_{5}\right)\right] \cdot \left[\left(-1-z_{1}\right) \cdots\left(-1-z_{5}\right)\right]}{\left[\left(i-z_{1}\right) \cdots\left(i-z_{5}\right)\right] \cdot \left[\left(-i-z_{1}\right) \cdots\left(-i-z_{5}\right)\right]} \\ & = \frac{\left(1^{5} + 2004 \cdot 1 - 1\right)\left(-1^{5} + 2004 \cdot (-1) - 1\right)}{\left(i^{5} + 2004 \cdot i - 1\right)\left(-i^{5} + 2004 \cdot (-i) - 1\right)} \\ & = \frac{(2004)(-2006)}{(-1 + 2005 i)(-1 - 2005 i)} \\ & = -\frac{2005^{2} - 1}{2005^{2} + 1} \\ & = -4020024 / 4020026 = -2010012 / 2010013 . \end{aligned}$$

Alternative Solution: In fact, we can construct the polynomial $P$ explicitly (up to multiplication by a constant). We write $P\left(z^{2}\right)$ as a polynomial in $z$; it must use only even powers of $z$ and be divisible by $z^{5} + 2004 z - 1$, so we are inspired to try a difference of squares,
$$P\left(z^{2}\right) = \left(z^{5} + 2004 z - 1\right)\left(z^{5} + 2004 z + 1\right) = \left(z^{5} + 2004 z\right)^{2} - 1^{2} = z^{2}\left(z^{4} + 2004\right)^{2} - 1,$$
giving
$$P(z) = z\left(z^{2} + 2004\right)^{2} - 1 .$$
Now plugging in $z = 1$ and $z = -1$ rapidly gives $\left(2005^{2} - 1\right) / \left(-2005^{2} - 1\right)$ as before.