Answer: 112

Small Hint:

Solve both inequalities for $k$

Big Hint:

Study the integers in the open interval $(\frac{6n}{7},\frac{7n}{8})$

Solution:

The inequalities are equivalent to $\frac{6n}{7}<k<\frac{7n}{8}.$ This interval has length $\frac{n}{56}.$ At $n=112$ it is $(96,98),$ containing only $97.$ For $n>112$ its length exceeds $2,$ so it contains at least two integers. Thus the largest possible $n$ is $112.$