Solution:
$2005^{1002} / 2004$ !
Consider the product
$$\left(\frac{1}{1}+\frac{1}{2004}\right)\left(\frac{1}{2}+\frac{1}{2003}\right)\left(\frac{1}{3}+\frac{1}{2002}\right) \cdots\left(\frac{1}{1002}+\frac{1}{1003}\right) .$$
This product expands to $2^{1002}$ terms, and each term gives the reciprocal of the value of a corresponding antipalindromic string of $P$ 's and $Q$ 's as follows: if we choose the term $1 / n$ for the $n$th factor, then our string has a $P$ in position $n$ and $Q$ in position $2005-n$; if we choose the term $1 /(2005-n)$, then we get a $Q$ in position $n$ and $P$ in position $2005-n$. Conversely, each antipalindromic string has its value represented by exactly one of our $2^{1002}$ terms. So the value of the product is the number we are looking for. But when we simplify this product, the $n$th factor becomes $1 / n+1 /(2005-n)=2005 / n(2005-n)$. Multiplying these together, we get 1002 factors of 2005 in the numerator and each integer from 1 to 2004 exactly once in the denominator, for a total of $2005^{1002}$ / $2004!$.