SAMC

Let us factor $30^{2010}=2^{2010}\cdot 3^{2010}\cdot 5^{2010}$.

Any divisor $d$ of $30^{2010}$ can be written as $d=2^a{3}^b{5}^c$ where $0\le a,b,c\le 2010$.

The product of two divisors $d_1=2^{a_1}{3}^{b_1}{5}^{c_1}$ and $d_2=2^{a_2}{3}^{b_2}{5}^{c_2}$ is a perfect square if and only if $a_1+a_2$, $b_1+b_2$, and $c_1+c_2$ are all even.

This is equivalent to $a_1\equiv a_2(\mathrm{mod}2)$, $b_1\equiv b_2(\mathrm{mod}2)$, $c_1\equiv c_2(\mathrm{mod}2)$.

Thus, for each divisor, consider the triple $(a\mathrm{mod}2,b\mathrm{mod}2,c\mathrm{mod}2)$. There are $2\times 2\times 2=8$ possible such triples.

If we select 9 divisors, by the pigeonhole principle, at least two of them must have the same triple. For these two divisors, the exponents of $2$, $3$, and $5$ have the same parity, so their sum is even for each prime, and thus their product is a perfect square.

Therefore, among any nine divisors of $30^{2010}$, there are two whose product is a perfect square.