Answer: (ii) Three.
(i) $2 = \frac{1}{1} + \frac{1}{1}$ and $\frac{5}{2} = \frac{2}{1} + \frac{1}{2}$ are good, thus $4 = 2 + 2$ and $5 = \frac{5}{2} + \frac{5}{2}$ are sums of two good numbers. For $n \ge 4$, we can write $n = 2(k+2)$ or $n = 2k+5$ with $k \ge 0$, thus $n$ is a sum of good numbers.

(ii) 57 is the sum of three good numbers:
$$\begin{aligned} 57 &= \left(\frac{52}{1} + \frac{1}{52}\right) + \left(\frac{45}{26} + \frac{26}{45}\right) + \left(\frac{20}{9} + \frac{9}{20}\right) \\ &= \left(\frac{75}{2} + \frac{2}{75}\right) + \left(\frac{25}{6} + \frac{6}{25}\right) + \left(\frac{15}{1} + \frac{1}{15}\right). \end{aligned}$$
Now we show three is the minimum.
First suppose $57 = \frac{a}{b} + \frac{b}{a}$ is good. We may assume $(a, b) = 1$. We have $a^2 + b^2 = 57ab \equiv 0 \pmod{3}$ and thus $a \equiv b \equiv 0 \pmod{3}$, and this contradicts $(a, b) = 1$.

Now suppose $57 = \frac{a}{b} + \frac{b}{a} + \frac{c}{d} + \frac{d}{c}$ with $(a,b) = (c,d) = 1$. Since $(a^2 + b^2, ab) = (c^2 + d^2, cd) = 1$ and
$$(a^2 + b^2)(cd) + (c^2 + d^2)(ab) = 57(ab)(cd),$$
we have $ab | cd$ and $cd | ab$. Thus $ab = cd$ and $a^2 + b^2 + c^2 + d^2 = 57ab = 57cd$. Moreover, we have
$$(a-b)^2 + (c+d)^2 = (a+b)^2 + (c-d)^2 = a^2 + b^2 + c^2 + d^2 \equiv 0 \pmod{3}.$$
It follows $a \equiv b \equiv 0 \pmod{3}$, and this contradicts $(a,b) = 1$. Hence at least three good numbers are needed to express 57 as their sum.