21.3.7 ** 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 and 120 are not removed). The remaining numbers form a new sequence: $a_{1}=1, a_{2}=2, a_{3}=5, a_{4}=7, \cdots$. Find $a_{1999}$.
Note that the least common multiple of $3$, $4$, and $5$ is 60, so we can first study the numbers that meet the requirements between 1 and 60.
Let $S=\{1,2, \cdots, 60\}, A_{1}=\{x|x \in S, 3| x\}, A_{2}=\{x|x \in S, 4| x\}$, and $A_{3}=\{x|x \in S, 5| x\}$. Then $|S|=60, |A_{1}|=20, |A_{2}|=15, |A_{3}|=12$, $|A_{1} \cap A_{2}|=5, |A_{1} \cap A_{3}|=4, |A_{2} \cap A_{3}|=3, |A_{1} \cap A_{2} \cap A_{3}|=1$. By the principle of inclusion-exclusion:
$$
|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}|+|A_{3}|=(60-20-15-12+5+4+3-1)+12=36,
$$

which means there are 36 numbers between 1 and 60 that are in the new sequence. Similarly, there are the same number of such numbers between 61 and 120, 121 and 180, and so on.
Since $1999=36 \times 55+19$, and it is easy to see that $a_{19}=31$, therefore $a_{1999}=60 \times 55+31=3331$.