Find the least integer $n$ such that the fractions are irreducible 
Find the least integer $n$ with the property that the fractions $$\dfrac{7}{n+9},\dfrac{8}{n+10},\dfrac{9}{n+11},\ldots,\dfrac{31}{n+33}$$ are irreducible.

In order for all the fractions to be irreducible, we must have $\gcd(n+k,k-2) = 1$ for $k = 9,10,\ldots,33$. We have $\gcd(n+k,k-2) = \gcd(n+2,k-2)$ and since $k-2 = 7,8,\ldots,31$, we must have $n+2$ relatively prime to $\text{lcm}(7,8,\ldots,31)$ for each $k$. Then take $n = 35$, which is the smallest $n$.

You just need to have $n\not \equiv -2 \bmod p$ for all prime $p$ not exceeding $31$.
In other words $n+2$ must not be divisible by any prime not exceeding $31$.
In other words we can take $n+2=37$ and so $n=35$ is the smallest options (because $377=35+2$ is the prime after $31$ (I am assuming you wanted $n$ to be positive, otherwise $n$ can be arbitrarily small).
