Bay Area Mathematical Olympiad

Solution:
We will show that this is possible exactly when $n$ is a perfect square larger than $1$. Suppose that we can reciprocate some of the fractions so that the resulting product is $1$. Let $r$ represent the product of the fractions that we will reciprocate and $t$ represent the product of the fractions that we will leave alone. Then $r \cdot t = n$ while $\frac{1}{r} \cdot t = 1$. Multiplying these equations shows that $n = t^{2}$, so $n$ is the square of a rational number, which means that it has to be a perfect square.

Now suppose that $n = a^{2}$ is a perfect square. Then we can reciprocate the first $a-1$ terms of the product to obtain
$$\left(\frac{1}{2}\right) \cdots \left(\frac{a-1}{a}\right) \left(\frac{a+1}{a}\right) \cdots \left(\frac{a^{2}}{a^{2}-1}\right) = \frac{1}{a} \cdot \frac{a^{2}}{a} = 1$$
demonstrating that modifying the product as desired is indeed possible for any perfect square.