7. 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 $\qquad$ .
7. 2048 Since $45^{2}=2025,46^{2}=2116$, so $2026=a_{2026-45}=a_{1981}, 2115=a_{2115-45}=a_{2070}$. And in the 90 terms from the 1981st term to the 2070th term, there are no perfect squares, and $1981+22=2003$, hence $a_{2003}=a_{1981}+22=2026+22=2048$.