2.3. Twenty thousand integers are marked, none of which divides any other. A number is called good if its square divides the product of some two other marked numbers. What is the maximum number of good numbers possible?

## Third Round
2.3. Answer: 2008. Example: numbers of the form $2^{k} 3^{2009-k}, k=0,1, \ldots, 2009$. It is clear that all numbers in this set, except for $2^{2010}$ and $3^{2010}$, are good. We will prove that there will always be at least two bad numbers. First, it is clear that the largest of our numbers $m$ is bad (the product of the other two is simply less than $m^{2}$). Let $n \neq m$ be another of our numbers, since $m$ is not divisible by $n$, there exists a prime divisor $p$ that enters the prime factorization of $n$ to a greater degree than in $m$. Choose from those of our numbers in which $p$ enters to the greatest degree, the largest number $k$. Note that $k$ is different from $m$. We will prove that the number $k$ is good. This follows from the fact that if $x y$ is divisible by $k^{2}$ for some of our numbers $x, y$, then, considering the degrees of entry of $p$ in $x, y, k$, we conclude that they are all equal, and then, due to the maximality of $k$, we have $x y < k^{2}$ - a contradiction.

## Third Round