A six-member company went fishing. Before departure, everyone predicted how many fish each would catch:

|  | who was tipped |  |  |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  |  | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ |  |
| who | $a$ | 9 | 4 | 8 | 3 | 1 | 6 |  |
| tipped | $b$ | 6 | 7 | 5 | 1 | 6 | 7 |  |
|  | $c$ | 6 | 9 | 16 | 5 | 10 | 3 |  |
|  | $d$ | 4 | 2 | 4 | 8 | 19 | 5 |  |
|  | $e$ | 9 | 5 | 7 | 9 | 12 | 9 |  |
|  | $f$ | 8 | 5 | 10 | 17 | 5 | 11 |  |

By evening, it turned out that everyone had predicted fewer fish for those who caught more than them, and more for themselves and those who caught fewer than them. No two of them caught the same number of fish, and everyone caught at least one fish. How many fish did each catch?
Each member's number of fish caught will be denoted by the corresponding capital letter, for example, $A$ for the number of fish caught by $a$. Our first observation is that $d$ could not have caught fewer fish than what $b$ guessed for him (which is 1), since everyone caught at least one fish. Thus, $d$ caught more fish than $b$, i.e., $D>B$. However, $b$ caught fewer fish than what $d$ guessed for him, so $B<2$, which means $B=1$. Therefore, $b$ caught fewer fish than everyone else, and thus guessed fewer fish for everyone, while everyone else guessed more fish for themselves, hence

$$
6<A<9, \quad 5<C<16, \quad 1<D<8, \quad 6<E<12, \quad 7<F<11.
$$

No one could have caught as many fish as were guessed for them, so $A$ cannot be 8, thus $A$ can only be 7. $c$ guessed fewer fish for $a$ (6) than what $a$ caught, so $C<A$, which, combined with (1), means $C=6$. From $D<8, D \neq A$, it follows that $d$ also caught fewer fish than $a$, so $d$ guessed more for himself, i.e., $D<3$, from which, given $B \neq D$, we have $D=2$. We now need to determine the number of fish caught by $e$ and $f$. From (1), $6<E$ and thus $c$ caught fewer fish than $e$, so $c$ also guessed fewer fish for $e$: $10<E$, from which $E=11$. Therefore, $e$ caught more fish than $f$ (since $F<11$), so $f$ guessed more for himself, i.e., $F<9$, from which $F=8$. In summary, the number of fish caught can only be as follows:

|  | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: |
| Number of fish caught: | 7 | 1 | 6 | 2 | 11 | 8 |

It remains to check that the rest of the data in the table is not in contradiction with our solution, but we leave this to the reader.

Note. If we do not specify that everyone must catch at least one fish, then the problem has additional solutions beyond the one given above, for example,

| $a$ | $b$ | $c$ | $d$ | $e$ | $f$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 7 | 3 | 6 | 0 | 11 | 8 |