Let $p_{1}, p_{2}, \ldots, p_{2005}$ be different prime numbers. Let $\mathrm{S}$ be a set of natural numbers which elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \ldots, p_{2005}$ and product of any two elements from $\mathrm{S}$ is not a perfect square.

What is the maximum number of elements in $\mathrm{S}$?
Let $a, b$ be two arbitrary numbers from $\mathrm{S}$. They can be written as

$$
a=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{2005}^{a_{2005}} \text { and } b=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{2005}^{\beta_{2005}}
$$

In order for the product of the elements $a$ and $b$ to be a square, all the sums of the corresponding exponents need to be even, from where we can conclude that for every $i$, $a_{i}$ and $\beta_{i}$ have the same parity. If we replace all exponents of $a$ and $b$ by their remainders modulo 2, then we get two numbers $a^{\prime}, b^{\prime}$ whose product is a perfect square if and only if $ab$ is a perfect square.

In order for the product $a^{\prime} b^{\prime}$ not to be a perfect square, at least one pair of the corresponding exponents modulo 2 need to be of opposite parity.

Since we form 2005 such pairs modulo 2, and each number in these pairs is 0 or 1, we conclude that we can obtain $2^{2005}$ distinct products none of which is a perfect square.

Now if we are given $2^{2005}+1$ numbers, thanks to Dirichlet's principle, there are at least two with the same sequence of modulo 2 exponents, thus giving a product equal to a square.

So, the maximal number of the elements of $S$ is $2^{2005}$.