We label the terms $x_{1}, x_{2}, x_{3}, \ldots, x_{2009}, x_{2010}$. Suppose that $S$ is the sum of the odd-numbered terms in the sequence; that is, $S=x_{1}+x_{3}+x_{5}+\cdots+x_{2007}+x_{2009}$. We know that the sum of all of the terms is 5307; that is, $x_{1}+x_{2}+x_{3}+\cdots+x_{2009}+x_{2010}=5307$. Next, we pair up the terms: each odd-numbered term with the following even-numbered term. That is, we pair the first term with the second, the third term with the fourth, and so on, until we pair the 2009th term with the 2010th term. There are 1005 such pairs. In each pair, the even-numbered term is one bigger than the odd-numbered term. That is, $x_{2}-x_{1}=1, x_{4}-x_{3}=1$, and so on. Therefore, the sum of the even-numbered terms is 1005 greater than the sum of the oddnumbered terms. Thus, the sum of the even-numbered terms is $S+1005$. Since the sum of all of the terms equals the sum of the odd-numbered terms plus the sum of the even-numbered terms, then $S+(S+1005)=5307$ or $2 S=4302$ or $S=2151$. Thus, the required sum is 2151.