Assume $n \in S$ is binary represented, i.e. $n \;=\; \sum_{i=1}^{k}c_{k}\cdot 2^{k}\;=\; c_{k}c_{k-1}\ldots c_{0}$ where $c_{i}\in \{0,1\}$ and $c_{k}= 1$. Then $[n/2] \;=\; c_{k}c_{k-1}\ldots c_{1}\in S.$ From this it follows inductively that $(1) \;\; [n/2^{i}] \;=\; c_{k}\ldots c_{i}\in S$ for every $i \in \{0,1,2, \ldots,k\}.$ Next let $N$ be a natural binary number. Now there exists an integer $M \in [2^{11}N, \: 2^{11}N+2002]$ such that $M \in S.$ Hence $M = 2^{11}N+R$ where $0 \leq R \leq 2002 < 2^{11},$ giving $(2) \;\; [M/2^{11}] \;=\; [N+{\textstyle \frac{R}{2^{11}}}] \;=\; N+[{\textstyle \frac{R}{2^{11}}}] \;=\; N \in S$ according to (1). Seeing that $N$ is an arbitrary natural number, (2) implies $S ={\bf N}.$

here, consider a abitrary $n\in N*$ . $n\in S$ if we found $2n$ or $2n+1$ in $S$. Similarly if we found any one of $4n,4n+1,4n+2,4n+3$in $S$ then also $n$.... Similarly , if we found any of $2^{2003}n,2^{2003}n+1,....,2^{2003}n+2002$ in $S$ then also $n$. but ${2^{2003}n,2^{2003}n+1,....,2^{2003}n+2002}$ are $2003$ consecutive natural numbers ,so at least on of them in $S$, hence the result follows,..