Solution:
Partition the interval $(0.00, 2.50]$ into 250 intervals each of size 0.01.
$$(0.00, 2.50] = (0.00, 0.01] \cup (0.01, 0.02] \cup (0.02, 0.03] \cup \dots \cup (2.49, 2.50].$$
Now consider all possible sets, $S$, we can choose from the given 10 positive integers with $|S| = 5$. Because each of the positive integers must be different, the smallest possible reciprocal sum of one of these sets is
$$1 + \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5} < 2.50.$$
Therefore each of these different sets has a reciprocal sum lying somewhere in the interval $(0.00, 2.50)$. The number of such sets $S$ is $\binom{10}{5} = 252$ but the number of intervals in our partition is only 250. By the pigeonhole principle there exists at least one interval, $(x, x + 0.01]$, and two distinct sets $S_{1}, S_{2}$ such that both reciprocal sums,
$$\sum_{s\in S_{1}}\frac{1}{s}\quad \mathrm{and}\quad \sum_{s\in S_{2}}\frac{1}{s}$$
lie in $(x, x + 0.01]$. The reciprocal sums of $S_{1}$ and $S_{2}$ have difference less than $1 / 100$ because they both lie in the interval $(x, x + 0.01]$. If $S_{1}$ and $S_{2}$ are disjoint then we can simply choose $A = S_{1}$ and $B = S_{2}$. Otherwise let $C = S_{1} \cap S_{2}$ be the intersection of $S_{1}$ and $S_{2}$ and then let $A = S_{1} \setminus C$ and let $B = S_{2} \setminus C$. The sets $A$ and $B$ are disjoint and equisized because $|S_{1}| = |S_{2}|$ and $S_{1} \neq S_{2}$.