I 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
Let the new sequence be $\left\{a_{n}\right\}$. Since $45^{2}=2025, 46^{2}=2116$. Therefore, $2026=a_{2126+45}=a_{1981}$. And $2115=a_{2115-45}=a_{2170}$. Thus, the terms between the 1981st and 2070th of the sequence are consecutive positive integers. Hence, $a_{20103}=a_{1981}+(2003-1981)=2026+22=2048$. The answer is (C).