Consecutive integers and divisibility

We have the relation $nu+(n-1)v=1$ for $u=1$ and $v=-1$. It is hence a Bezout relation between $n$ and $n-1$ that proves that these integers have gcd $1$.
Hence $n-1$ divides $n$ if, and only if, $n-1=1$ (so $n=2$) of $n-1=n$, which is not possible.

The answer is thus $2$.