Taiwan IMO Selection Camp

Argue indirectly. Agree, as usual, that the empty sum is $0$ to consider rationals in $[0, 1)$; adjoining $0$ causes no harm, since $\sum_{x \in F} 1/x = 0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0, 1)$, let $F_r$ be the unique finite subset of $S$ such that
$$\sum_{x \in F_r} \frac{1}{x} = r.$$
The argument hinges on the lemma below.

Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0, 1)$ such that $q - r = 1/x$, then $x$ is a member of $F_q$ if and only if it is not one of $F_r$.

Proof. If $x$ is a member of $F_q$, then
$$\sum_{y \in F_q \setminus \{x\}} \frac{1}{y} = \sum_{y \in F_q} \frac{1}{y} - \frac{1}{x} = q - \frac{1}{x} = \sum_{y \in F_r} \frac{1}{y},$$
so $F_r = F_q \setminus \{x\}$, and $x$ is not a member of $F_r$. Conversely, if $x$ is not a member of $F_r$, then
$$\sum_{y \in F_r \cup \{x\}} \frac{1}{y} = \sum_{y \in F_r} \frac{1}{y} + \frac{1}{x} = r + \frac{1}{x} = q = \sum_{y \in F_q} \frac{1}{y},$$
so $F_q = F_r \cup \{x\}$, and $x$ is a member of $F_q$. □

Consider now an element $x$ of $S$ and a positive rational $r < 1$. Let $n = [rx]$ and consider the sets $F_{r-k/x}$, $k = 1, \dots, n$. Since
$$0 \le r - \frac{n}{x} < \frac{1}{x},$$
the set $F_{r-n/x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2k)/x}$ do not contain $x$, whereas the $F_{r-(n-2k-1)/x}$ do. Consequently, $x$ is a member of $F_r$ if and only if $n$ is odd.

Finally, consider $F_{2/3}$. By the preceding, $[2x/3]$ is odd for each $x$ in $F_{2/3}$, so $2x/3$ is not integral. Since $F_{2/3}$ is finite, there exists a positive rational $\epsilon$ such that $[(2/3 - \epsilon)x] = [2x/3]$ for all $x$ in $F_{2/3}$. This implies that $F_{2/3}$ is a subset of $F_{2/3-\epsilon}$ which is impossible.

A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \setminus \{1\}$. We may and will therefore assume that $S$ consists of integers greater than $1$. Label the elements of $S$ increasing $x_1 < x_2 < \dots$, where $x_1 \ge 2$.

We first show that $S$ satisfies (2) if $x_{n+1} \ge 2x_n$ for all $n$. In this case, $x_n \ge 2^{n-1}x_1$ for all $n$, so
$$s = \sum_{n \ge 1} \frac{1}{x_n} \le \sum_{n \ge 1} \frac{1}{2^{n-1}x_1} = \frac{2}{x_1}.$$
If $x_1 \ge 3$, or $x_1 = 2$ and $x_{n+1} > 2x_n$ for some $n$, then
$$\sum_{x \in F} \frac{1}{x} < s < 1 \text{ for every finite subset } F \text{ of } S,$$
so $S$ satisfies (2); and if $x_1 = 2$ and $x_{n+1} = 2x_n$ for all $n$, that is, $x_n = 2^n$ for all $n$, then every finite subset $F$ of $S$ consists of powers of $2$, so
$$\sum_{x \in F} \frac{1}{x} \ne \frac{1}{3}$$
and again $S$ satisfies (2).

Finally, we deal with the case where $x_{n+1} < 2x_n$ for some $n$. Consider the positive rational
$$r = \frac{1}{x_n} - \frac{1}{x_{n+1}} < \frac{1}{x_{n+1}}.$$
If $r = \sum_{x \in F} 1/x$ for no finite subset $F$ of $S$, then $S$ satisfies (2).

We now assume that $r = \sum_{x \in F_0} 1/x$ for some finite subset $F_0$ of $S$, and show that $S$ satisfies (1). Since
$$\sum_{x \in F_0} \frac{1}{x} = r < \frac{1}{x_{n+1}},$$
it follows that $x_{n+1}$ is not a member of $F_0$, so
$$\sum_{x \in F_0 \cup \{x_{n+1}\}} = \sum_{x \in F_0} \frac{1}{x} + \frac{1}{x_{n+1}} = r + \frac{1}{x_{n+1}} = \frac{1}{x_n}.$$
Consequently, $F = F_0 \cup \{x_{n+1}\}$ and $G = \{x_n\}$ are distinct finite subsets of $S$ such that
$$\sum_{x \in F} \frac{1}{x} = \sum_{x \in G} \frac{1}{x},$$
and $S$ satisfies (1).