1. There are three consecutive positive integers, the sum of whose reciprocals is $\frac{47}{60}$. Find these three numbers.
Let the three numbers be $x, x+1, x+2$, then we have
$$
\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}=\frac{47}{60} .
$$

Eliminating the denominators and rearranging, we get
$$
47 x^{3}-180 x^{2}-219 x-26=0 \text {. }
$$

This equation is generally not solved in high school, and even if it can be solved, the calculation is very complicated. Based on the characteristics of the original equation, it can be solved using the method of inequalities, which is very simple.
From the original equation, we can easily obtain
$$
\left\{\begin{array}{l}
\frac{1}{x}>\frac{47}{60 \times 3}, \\
\frac{1}{x+2}<\frac{8}{47}
\end{array} .\right.
$$

That is, $x=3$, which, upon verification, satisfies the original equation.
Therefore, the three numbers are $3,4,5$.
In general, if there are $n$ consecutive positive integers, and the sum of their reciprocals is a constant $k$, to find these $n$ positive integers, we can set the $n$ numbers to be $x, x+1, x+2, \cdots, x+n-1$. According to the problem, we have $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\cdots+\frac{1}{x+n-1}=k$. From the equation, we get
$$
\left\{\begin{array}{l}
\frac{1}{x}>\frac{k}{n}, \\
\frac{1}{x+n-1}<\frac{k}{n}
\end{array} \Rightarrow 1-n+\frac{n}{k}<x<\frac{n}{k} .\right.
$$

Solving this system of inequalities, find the positive integer solution for $x$, and then substitute it back into the original equation for verification, which will yield the solution to the original problem.