8. The positive integer $n$ ensures that in every $n$-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (which can be the same), whose sum is a positive integer power of 2. Then the minimum value of $n$ is $\qquad$
8.1003.

First, if $n=1002$, take
$$
\begin{array}{l}
A=\{5,6,7\}, B=\{17,18, \cdots, 24\}, \\
C=\{33,34, \cdots, 39\}, \\
D=\{1025,1026, \cdots, 2008\} .
\end{array}
$$

Then $S=A \cup B \cup C \cup D$.
Thus, $|S|=3+8+7+984=1002$.
It is easy to verify that the sum of any two (possibly the same) elements in $S$ is not a power of 2.

Second, prove: In every 1003-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (possibly the same) whose sum is a power of 2.

In fact, divide the set $\{1,2, \cdots, 2008\}$ into the union of 1003 pairwise disjoint subsets as follows:
$$
\begin{array}{l}
\{1,7\},\{2,6\},\{3,5\},\{8,24\},\{9,23\}, \\
\{10,22\}, \cdots,\{15,17\},\{25,39\}, \cdots, \\
\{31,33\},\{40,2008\},\{41,2007\}, \\
\{42,2006\}, \cdots,\{2003,1025\}, \\
\{4,16,32,1024\} .
\end{array}
$$

Each of the above sets has 2 elements (possibly the same) whose sum is a power of 2. Therefore, every 1003-element subset of $\{1,2, \cdots, 2008\}$ either contains one element from $\{4,16,32,1024\}$ or contains one of the first 1002 sets. Thus, there are 2 elements (possibly the same) whose sum is a power of 2.
In summary, the minimum value of $n$ is 1003.