13. From the sequence of positive integers $1,2,3,4, \cdots$, remove the multiples of 3 and 4, but retain all multiples of 5 (for example, $15,20,60,120$ are not removed). After the removal, the remaining numbers form a sequence in their original order: $a_{1}=1, a_{2}=$ $2, a_{3}=5, a_{4}=7, \cdots$. Find $a_{2003}$.
13. Since the least common multiple of $3,4,5$ is 60, let $S=\{1,2, \cdots, 60\}, A_{i}=\{k \mid k \in S$ and $k$ is divisible by $i\}(i=3,4,5)$, then the set of numbers in $S$ that are not divided is $\left(\bar{A}_{3} \cap \bar{A}_{4} \cap \bar{A}_{5}\right) \cup A_{5}$. By the principle of inclusion-exclusion, we can calculate $\left(\bar{A}_{3} \cap \bar{A}_{4} \cap \bar{A}_{5}\right) \cup \left.A_{5}\right\}=36$. Let $P=\left\{a_{1}, a_{2}, \cdots, a_{36}\right\}$, and it can be proven that the terms in the sequence $\left\{a_{n}\right\}$ consist exactly of numbers of the form $k \cdot 60+r$ (where $k$ is a non-negative integer, $r \in P$). Also, $2003=36 \times 55+23$, so $a_{2003}=60 \times 55+a_{23}$, and $a_{36}=60, a_{35}=59, a_{34}=58$, $a_{33}=55, a_{32}=53, a_{31}=50, a_{30}=49, a_{29}=47, a_{28}=46, a_{27}=45, a_{26}=43, a_{25}=41, a_{24}=40, a_{23}=38$, hence $a_{2003}=60 \times 55+38=3338$