Find all integers $n \geqslant 1$ such that $n$ divides $2^{n}-1$.
First, $n=1$ works. Conversely, suppose we have a solution $n \geqslant 2$. Let $p$ be the smallest prime factor of $n$. Since $2^{n}-1$ is odd, $n$ is also odd, hence $p \geqslant 3$.

Let $\omega$ be the multiplicative order of 2 modulo $p$. Since $\omega$ divides $n$ and divides $p-1$, it also divides $n \wedge(p-1)=1$, so $p$ divides $2^{1}-1=1$.

This impossibility shows that $n=1$ is the only solution.