It's easy to show that the maximum value of $ n$ is $ 1958$, and can be attained with $ S = \{45,46,\ldots, 2002\}$. Since $ 45^2 > 2002$, obviously this set $ S$ fulfills the hypothesis. Suppose we could find a set $ S_0$ with 1959 elements. Let's say the number of elements of $ S_0$ less than or equal to 44 is $ a$, and greater than or equal to $ 45$ is $ b$. Then $ a + b = 1959$, $ 0\leq a\leq 43$ (obviously $ 1 \notin S$), and $ 1959 - 43 = 1916\leq b \leq 1958$. First let's note that none of the numbers $ 2,3,4,5,6$ can be in the set $ S_0$. Indeed, if any of those numbers, say $ k \in\{2,3,4,5,6\}$, would be in such a set $ S_0$, then we can only have in $ S_0$ one of the elements of each of the following pairs \[ (45, 45k), (46, 46k), \ldots, (89,89k) , \] so $ b\leq 1958 - (89 - 45 + 1) = 1913$, contradiction. Why do we need this result? Because starting $ 7$, for each element $ x$ in $ S_0$, $ x^2 > 45$, therefore for each element $ x\leq 44$ in $ S_0$, its square must not be in $ S_0$, so $ b \leq 1958 - a$. However $ a + b = 1959$, which is a contradiction.

44×44=1936

yunxiu wrote: 44×44=1936 Oops The proof is still valid. I've modified it to correct this mistake

Peter wrote: Determine the largest positive integer $n$ for which there exists a set $S$ with exactly $n$ numbers such that each member in $S$ is a positive integer not exceeding $2002$, if $a,b\in S$ (not necessarily different), then $ab\not\in S$. We can prove $n\le 1958$ by considering 44 triples $(44, 45, 44\times 45), (43, 46, 43\times 46),\ldots,(1,88,1\times 88)$. In each triple $(44-i,45+i,(44-i)(45+i)), 0\le i\le 43$, there are at most two numbers belong to $S$. So, $|S|\le 2002-44=1958$.