The claim obviously hold when $n=1$. When $n > 1$, let's consider the following $n$ numbers: $$1, 2, \cdots, 2^{n-1}$$

They are all powers of $2$, but $n$ is odd, therefore none of them is a multiple of $n$. It follows that at least two of them are congruent modulo $n$. Let these two numbers be $a^i$ and $2^j$, where $0 \le i < j \le n-1$. Now we have $$2^i\equiv 2^j\pmod{n} \implies 2^i(2^{j-i}-1)\equiv 0\pmod{n}$$

Again, $2^i\not\equiv 0\pmod{n}$. Therefore $2^{i-j}\equiv 0\pmod{n}$ where $1 \le i - j \le n-1$.