Solution:
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 $a'$, $b'$ whose product is a perfect square if and only if $ab$ is a perfect square.

In order for the product $a' b'$ 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 $1$ or $2$, 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 $2^{2005}$.