Which is the time between 2 and 3 o'clock, for which there is a time between 6 and 7 o'clock such that the positions of the clock hands - with the hands swapped - are the same.
I. solution: It is useful to measure the distance traveled by the hands on the clock face in terms of the 60th part of a full circle (the distance traveled by the minute hand in one minute) or the 12th part (the distance traveled by the hour hand in one hour).

At the hand position between 2 and 3 o'clock, let the distance traveled by the hour hand from 2 o'clock, measured in 60th parts of a full circle, be $x$. Then the hour hand is $10 + x$ units away from the 12, and the minute hand is $12x$ units away.

If after 6 o'clock the hour hand has moved $y$ units to reach the same position with the hands swapped, then in the second position the hour hand is $30 + y$ units away from the 12, and the minute hand is $12y$ units away. According to the condition,

$$
\begin{aligned}
10 + x & = 12y \\
12x & = 30 + y
\end{aligned}
$$

must hold. From this,

$$
x = \frac{370}{143} = 2 \frac{84}{143}, \quad y = \frac{150}{143} = 1 \frac{7}{143}
$$

Thus, the sought time is

$$
2 \text{ o'clock } 31 \frac{7}{143} \text{ minutes} (\approx 31 \text{ min } 2.9 \text{ sec}).
$$

The corresponding time between 6 and 7 o'clock is

$$
6 \text{ o'clock } 12 \frac{84}{143} \text{ minutes} (\approx 12 \text{ min } 35.2 \text{ sec}).
$$

II. solution: We can also derive equations for the unknowns from the fact that at any time, the hour hand divides the arc between two adjacent hour marks in the same ratio as the arcs from the 12 in the clockwise and counterclockwise directions to the minute hand.

Let the unit be the 12th part of a full circle, and let one of the hands be $x$ units after the 2 and the other $y$ units after the 6 at the desired hand position. Then, for the time between 2 and 3 o'clock,

$$
\frac{x}{1-x} = \frac{6+y}{6-y}
$$

and for the hand position between 6 and 7 o'clock,

$$
\frac{y}{1-y} = \frac{2+x}{10-x}
$$

From these,

$$
12x = 6 + y, \quad 12y = 2 + x
$$

If we also take into account that the units here are fifths of the $x$ and $y$ values in the first solution, we can see that we essentially got back the system of equations in the first solution.

III. solution: The problem can also be solved with one unknown. Using the notation of the first solution, the positions of the hour and minute hands between 2 and 3 o'clock are now given by the rotations $10 + x$ and $12x$. If the hour hand is now at $12x$ between 6 and 7, then the position of the minute hand is indicated by the rotation $12(12x - 30)$, so

$$
10 + x = 12(12x - 30)
$$

Here, we essentially eliminated the unknown $y(=12x - 30)$ from the system of equations in the first solution through reasoning.

Note: Many chose the following method to determine the position of the hands (again, measuring the rotations in 12th parts of a full circle): At the hand position between 2 and 3 o'clock, the minute hand is between 6 and 7. This means that the hour hand can be at least $\frac{6}{12}$ and at most $\frac{7}{12}$ away from the 2. Then, between 6 and 7 o'clock, the minute hand is between $\frac{30}{12}$ and $\frac{31}{12}$ from the 12, so the hour hand must be at least $\frac{30}{144}$ and at most $\frac{31}{144}$ after the 6. This means the minute hand is also between $\frac{894}{144}$ and $\frac{895}{144}$ from the 12, or at least $\frac{894}{12^3}$ and at most $\frac{895}{12^3}$ from the 2. The positions of the hands can be narrowed down further in this manner. For example, $\frac{1}{12^3} = \frac{1}{1728}$ of a unit (or a part of 5 minutes) is less than 0.2 seconds, so the obtained values already provide a good approximation. This step-by-step approximation method can also be used to determine the exact value, but its thorough exploration would require clarifying significant new concepts, which would not be proportional to the difficulty of the problem.

Remark: Many contestants shied away from the common fractions that provide the exact result and instead used decimal fractions to calculate imprecisely.