Let the 3 smallest fishes weight a, b and c kilograms.  Then  

    a + b + c = 25 kg     (1)


    Notice, that from equation (1), the weight at least one of the fishes {a,b,c}  MUST BE MORE than  8 kilograms
    (otherwise, their total weight would be not more than 24 kg).

    Let call this fish M8 ("more than 8 kilograms"):  M8 > 8 kg.    <<<---=== note the strong inequality (!)



Let the 3 largest fishes weight x, y and z kilograms.  Then  

    x + y + z = 35 kg     (2)



The sum (1) plus sum (2) is 25 + 35 = 60 kilograms, which  is 40 kilograms less than 100 kilograms.


Hence, it should be at least 4 or more other fishes, distinct from a, b, c, x, y and z, to balance this difference of 40 kilograms.


    Now, if the set of these distinct fishes contain 5 or more fishes, then the weight of at least one of these 5 or more distinct fishes 
    must be LESS than OR equal to 8 kilograms

    (otherwise, the total weight of these 5 distinct fishes would be more than 40 kilograms).

    Let call this fish L8 ("less or equal to 8 kilograms"):  L8 <= 8.



Then, replacing M8 by L8 in the set {a,b,c}, we would obtain the new set of 3 fishes weighing in total LESS THAN 25 kilograms.


It gives us a CONTRADICTION with the statement that the three smallest fishes weight 25 kilograms.


The contradiction means that the set of distinct fishes consists of EXACTLY 4 fishes.


Hence, the total number of fishes is 3 + 4 + 3 = 10.      ANSWER