Problem 562

Explanation: We solve a more general problem, finding the general term formula of the new sequence $\left\{a_{n}\right\}$.

Since $(k+1)^{2}-k^{2}-1=2 k, k \in \mathbf{Z}_{+}$, in the sequence of positive integers $1,2,3, \cdots$, there are $2 k$ numbers between $k^{2}$ and $(k+1)^{2}$, all of which are terms of $\left\{a_{n}\right\}$. All these numbers, arranged in ascending order, form the sequence $\left\{a_{n}\right\}$. We group the sequence $\left\{a_{n}\right\}$ as follows:
$$\begin{array}{c} (2,3),(5,6,7,8),(10,11,12,13,14,15), \\ \cdots,\left(k^{2}+1, k^{2}+2, \cdots, k^{2}+2 k\right), \cdots, \end{array}$$

where the $k$-th group has $2 k$ numbers.
If $a_{n}$ is in the $k$-th group, then
$$a_{n}=k^{2}+n-[2+4+\cdots+2(k-1)]=n+k,$$

and
\begin{array}{l} {[2+4+\cdots+2(k-1)]+1} \\ \leqslant n0 . \end{array}\right.

Noting that $k>0$, we solve to get
$$-\frac{1}{2}+\sqrt{n-\frac{3}{4}}<k \leqslant \frac{1}{2}+\sqrt{n-\frac{3}{4}} \text {. }$$

Since $k \in \mathbf{Z}_{+}$, and
$$\frac{1}{2}+\sqrt{n-\frac{3}{4}}-\left(-\frac{1}{2}+\sqrt{n-\frac{3}{4}}\right)=1 \text {, }$$

we have $k=\left[\frac{1}{2}+\sqrt{n-\frac{3}{4}}\right]$.
Therefore, $a_{n}=n+\left[\frac{1}{2}+\sqrt{n-\frac{3}{4}}\right]$.
When $n=2003$,
$$a_{20 B}=2003+\left[\sqrt{2003-\frac{3}{4}}+\frac{1}{2}\right]=2048 \text {. }$$

Note: This problem has appeared in various forms in different mathematics competitions, such as
(1) Prove that in the sequence of positive integers, after removing all perfect squares, the $n$-th term is equal to $n+(n)$, where $(n)$ represents the integer closest to $\sqrt{n}$.
(27th Putnam Mathematical Competition $\mathrm{A}-4$ )
(2) Prove that if the positive integer $A$ is not a perfect square, then there exists a positive integer $n$ such that
$$A=\left[n+\sqrt{n}+\frac{1}{2}\right] \text {. }$$
(1992, St. Petersburg Selection Exam)