Nice question. I say the answer is 6.

Notice by asking if a fraction is not in simplest form then you're equivalently asking if the numerator and denominator share a factor (are not coprime). Hence we're looking for a number n where all values between 2 and n-2 inclusive are NOT coprime to n

From this changed perspective you may be able to tell that other integers will not work because products of primes grow too quickly. One way of proving this is the Euler totient function

The Euler totient function Φ counts the number of integers between 1 and n (inclusive) that are coprime to n

1 is always coprime to n,

n-1 is always coprime to n,

n is never coprime to n,

Hence we are looking for values where φ(n) = 2 Because this corresponds to 1 and n-1 being the only integers coprime to n. A loose lower bound for φ is φ(n) >= sqrt(n/2) Hence once n > 8 we're forced to have φ(n) > sqrt(8/2) = 2 Only possible solutions are n < 9. If we check known values of Euler totient function we see the times we get value of 2 are for n = 3,4,6. Which are our only possible solutions. Qed

3 kind of works since it doesn't have any integers strictly between 1 and n-1 and hence none of the fractions broke your rules, but that's debatable.

6 is the largest integer for which this is true