At most how many numbers can we select from the first 1983 positive integers so that the product of any two selected numbers does not appear among the selected numbers?
Clearly, we get an appropriate set of numbers if the product of any two of the selected numbers is greater than 1983. Since $1983 < 45 \cdot 46$, selecting the numbers 45, 46, ..., 1982, 1983 results in an appropriate set of numbers. This means that we can select 1939 numbers that meet the conditions.

If $a \cdot b \leq 1983$, then at most two of $a$, $b$, and $a b$ can be among the selected numbers. Let's try the products of the form $(44-i)(45+i)=1980-i-i^{2}$! Denote the factors of this product by $a_{i}, b_{i}$, and the product itself by $c_{i}$ for $i=1, \ldots, 42$.

It immediately follows that $c_{i+1}<c_{i}$, and also $b_{42}=87<194=c_{42}$, so the following inequalities hold among the numbers $a_{i}, b_{i}$, and $c_{i}$: $2=a_{42}<a_{41}<\ldots<a_{0}<b_{0}<\ldots<b_{42}<c_{42}<c_{41}<\ldots<c_{0}=1980$.

From this, out of the $3 \cdot 43=129$ numbers, at most $2 \cdot 43=86$ can be among the selected numbers. From the remaining $1983-129=1854$ numbers, it is not worth selecting the number 1, as it would exclude every other number. Thus, $1854 + 86 = 1940$ is the maximum number of numbers that can be selected to meet the conditions.

Therefore, we have found that the maximum set of numbers that meets the conditions consists of 1939 elements.