Solution. Let the standing time be denoted by $x$ for the outbound trip and $y$ for the return trip. Thus, according to the text, the postman traveled for $3x$ time on the return trip and $4y$ time on the outbound trip. Summarizing this in a table:

|  | Outbound | Return |
| :--- | :---: | :---: |
| stand | $x$ | $y$ |
| travel | $4 y$ | $3 x$ |

Furthermore,

$$
x+4 y=y+3 x, \quad \text { i.e., } \quad 3 y=2 x
$$

According to this: $3 x=\frac{9}{2} y$. The ratio of the average speeds during travel is:

$$
\frac{\text { Outbound }}{\text { Return }}=\frac{\frac{s}{4 y}}{\frac{s}{3 x}}=\frac{\frac{s}{4 y}}{\frac{s}{\frac{9}{2}}}=\frac{\frac{9}{2}}{4}=\frac{9}{8}
$$
Pali, when delivering letters on a long street, first delivered on the odd side of the street in one direction, then on the even side in the opposite direction. On the way there, he spent three times as long at the mailboxes as he did walking back. On the way back, he spent a quarter of the time at the mailboxes compared to the time he spent walking there. In the end, it turned out that the delivery took the same amount of time on both sides of the street. How does the average walking speed (excluding stops) on the way there compare to the average walking speed on the way back?