8. The number of non-empty subsets of the set $\{1,2, \cdots, 2009\}$ whose elements sum to an odd number is $\qquad$.
8. $2^{2008}$.

Solution 1 Let
$$
f(x)=(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{2009}\right) \text {. }
$$

Then the answer required in the problem is the sum of the coefficients of the odd powers of $x$ in the expansion of $f(x)$.
Thus, the answer is
$$
\frac{f(1)-f(-1)}{2}=\frac{2^{2009}-0}{2}=2^{2008} \text {. }
$$

Solution 2 Consider the subsets $A$ of the set $\{1,2, \cdots, 2009\}$ that do not contain 2009. If the sum of the elements in $A$ is even, then add 2009; otherwise, do not add it.

Hence, there are $2^{2008}$ non-empty subsets with an odd sum of elements.