Solution:

Let $R_{n}$ be the set of all possible resistances using exactly $n$ $1$-ohm circuit segments (without shorting any of them), then we get
$$R_{n} = \bigcup_{i=1}^{n-1} \left( \{ a+b \mid a \in R_{i},\ b \in R_{n-i} \} \cup \left\{ \left. \frac{a b}{a+b} \ \right|\ a \in R_{i},\ b \in R_{n-i} \right\} \right)$$
starting with $R_{1} = \{1\}$, we get:
$$\begin{aligned} R_{2} & = \left\{ \frac{1}{2},\ 2 \right\} \\ R_{3} & = \left\{ \frac{1}{3},\ \frac{2}{3},\ \frac{3}{2},\ 3 \right\} \\ R_{4} & = \left\{ \frac{1}{4},\ \frac{2}{5},\ \frac{3}{5},\ \frac{3}{4},\ 1,\ \frac{4}{3},\ \frac{5}{3},\ \frac{5}{2},\ 4 \right\} \end{aligned}$$
Their union is the set of all possible effective resistances we can get, which contains $2+4+9=15$ values. (Note that $R_{1} \subset R_{4}$ and the sets $R_{2}$, $R_{3}$, $R_{4}$ are disjoint.)