Consider this: 23/1,14/2,15/3,24/4,25/5,12/6,21/7,16/8,18/9,20/10,22/11,26/13 and non integer 19/17

Proof that there is no more than 12, i.e. 13. Suppose we have 13 fractions, all integers. We have two primes 17 and 19 and we have no other multiples of 17 and 19, but we have only one 1. Hence, we cannot have 13 fractions and since Nymoldin's presented configuration with 12 fractions works, our answer is $\boxed{12}$.

Is there any generalized form for {1,2,3,...n}?

We claim that the answer is $\boxed{12}$, it is impossible to be $14$, so we test $13.$ We have the primes $17,19,23$ (13 can be 26/2) so one of them we can use $1$ on 23 but we have 17 and 19, so it is impossible for $13.$ Now we take $12$, we have 23/1 14/2 15/3 24/4 25/5 12/6 21/7 16/8 18/9 20/10 22/11 26/13

First we must write primes till 26. They are;

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${2,3,5,7,11,13,17,19,23}$ Then match primes which are bigger than 7.

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$(23,1), (24,4), (25,5), (26,13), (22,11) , (16, 8) ,(20, 10), (21, 7) , (15,3) , (18, 9), (14, 2) , (12, 6)$ There are no pairs for 17 and 19. So, the answer is 12.

lian_the_noob12 wrote: Is there any generalized form for {1,2,3,...n}? look for primes $p>\lfloor \frac{n}{2}\rfloor$ maybe this would work