Coprime remainders

Residues. If $a+i\equiv a+j\pmod n$ with $0\le i<j<n$, then $n\mid(j-i)$, but $0<j-i<n$, impossible. So the residues are $0,1,\ldots,n-1$ in some order.

Exactly one divisible. Exactly one residue is $0$, so exactly one integer in the sequence is divisible by $n$.

Conclusion. The statement holds.

Application. Among any $n$ consecutive integers, one is a multiple of each divisor $d\mid n$.