$7$ fishermen caught exactly 100 fish and no two had caught the same number of fish. Then there are three who have together captured at least 50 fish.

Let's work with the lowest four numbers instead of the other suggestions.

Supposing there is a counterexample, then the lowest four must add to at least $51$ (else the highest three add to $50$ or more).

Since $14+13+12+11=50$ the lowest four numbers would have to include one number at least equal to $15$ to get a total as big as $51$.

Then the greatest three numbers must be at least $16+17+18=51$, which is a contradiction to the assumption that there exists a counterexample.

The examples $18+17+15+14+13+12+11=100$ and $19+16+15+14+13+12+11=100$ show that the bound is tight.

If the maximum number of fish caught is $m$, then the total number of fish caught is no more than $m+(m-1)+...+(m-6)$. So there is one fisherman that caught at least 18 fish. Repeat this process for the second and third highest number of fish caught and you should be good.

I should add that this is a common proof technique in combinatorics and graph theory. To show that something with a certain property exists, choose the "extremal" such something, and prove that property holds for the extremal object. For instance, to show in a graph where each vertex has degree at least $d$ there is a path of length at least $d$, and one proof starts by simply showing a maximal path has length at least $d$.

I think I have a solution. First note that if $r_4 \geq 15$ then we have:

$r_5 \geq 16$

$r_6 \geq 17$

$r_8 \geq 18$

so $r_5 + r_6 + r_7 \geq 16 + 17 +18 = 51$ which is impossible.

Therefore $r_4 < 15$

Also note that if $r_4 \leq 14$ then:

$r_3 \leq 13$

$r_2 \leq 12$

$r_1 \leq 11$

thus $r_1 + r_2 + r_3 + r_4 \leq 50$ which implies that $r_5 + r_6 + r_7 \geq 50$ so that cannot be. Thus we have $r_4 > 15$ which is a contradiction.