I. Multiple Choice Questions (6 points per question, total 36 points)
1. 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
(C)
1. 【Analysis and Solution】Solution 1 Notice that $45^{2}=2025,46^{2}=2116$,
$$
\therefore 2026=a_{2026-45}=a_{1981}, 2115=a_{2115-45}=a_{2077} \text {. }
$$

Moreover, there are no perfect squares in the 90 terms from the 1981st term to the 2070th term. Also, $1981+22=2003$, $\therefore a_{2003}=a_{1981}+22=2026+22=2048$. Therefore, the answer is ( $\left.\mathrm{C}\right)$.

Solution 2 Group the new sequence according to the rule that the $k$-th group contains $2k$ numbers: $(2,3),(5,6,7,8),(10,11$, $12,13,14,15), \cdots$. Suppose the 2003rd term of the new sequence is in the $n$-th group, then $2+4+6+\cdots 2n \leqslant 2003$, i.e., $n(n+1)$
$\leqslant 2003$.
Thus, $n \leqslant 44$.
Therefore, the 2003rd term of the new sequence is
$$
44 \times 45+45+23=2048 \text {. }
$$