7. Let $A$ be the set of all positive integers not exceeding 2009, i.e., $A=\{1,2, \cdots, 2009\}$, and let $L \subseteq A$, where the difference between any two distinct elements of $L$ is not equal to 4. Then the maximum possible number of elements in the set $L$ is
7. 1005.

Divide the set $A$ into the following 1005 subsets:
$$
\begin{array}{l} 
A_{4 k+i}=\{8 k+i, 8 k+i+4\}(i=1,2,3,4 ; \\
k=0,1, \cdots, 250), \\
A_{1005}=\{2009\} .
\end{array}
$$

If the number of elements in $L$ is greater than 1005, then at least one of the first 1004 subsets is a subset of $L$, meaning there exist two elements in $L$ whose difference is 4.

Furthermore, the set $L$ formed by taking one element from each of the first 1005 subsets satisfies the condition.

Therefore, the maximum possible number of elements in the set $L$ is 1005.