Example 9 (2003 National High School Competition Question) Remove all perfect squares from the sequence of positive integers $1,2,3 \cdots$, to get a new sequence. The 2003rd term of this new sequence is ( ).
A. 2046
B. 2047
C. 2048
D. 2049
Solution: Choose C. Reason: Consider the square number greater than 2003 and closest to it, which is $45^{2}=2025$ (since $44^{2}=1936$). At this point, we have $2026=a_{2026-45}=a_{1981}$.

Thus, we also need to consider $46^{2}=2116$, at which point, $2115=a_{2115-45}=a_{2077}$. Since there are no perfect squares among the 90 terms from the 1981st term to the 2070th term. And $1981+22=2003$, then $a_{2003}=a_{1981}+22=2026+22=2048$.