4. Remove all perfect squares from the sequence of positive integers $1,2,3, \cdots$, to obtain a new sequence. The 2003rd term of this new sequence is
A. 2046
B. 2047
C. 2048
D. 2049
4. Let $\mathrm{C}$ denote the sequence obtained by removing all perfect squares from the sequence of positive integers $1,2,3, \cdots$, resulting in a new sequence $\left\{a_{n}\right\}$. Noting that $45^{2}=$ $2025,46^{2}=2116$, we have $2026=a_{2026-45}=a_{1981}, 2115=a_{2115-45}=a_{2070}$. Moreover, there are no perfect squares in the 90 terms from the 1981st term to the 2070th term. Since $1981+22=2003$, it follows that $a_{2003}=a_{1981}+22=2048$.