Find all odd integers $n \geqslant 1$ such that $n$ divides $3^{n}+1$.

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Let $n>1$ such that $n$ divides $3^{n}+1$. Another way to say this is that $3^{n} \equiv-1(\bmod n)$, so we are dealing with a purely multiplicative problem. We would like to use Fermat's theorem, but it is difficult to exploit because we have poor control over $\varphi(n)$. Let $p$ be a prime factor of $n$, which is odd. We have $3^{2 n} \equiv 1(\bmod p)$. Let $\omega$ be the order of 3 modulo $p$. Then $\omega$ divides $2 n$. On the other hand, by Fermat's little theorem, $3^{p-1} \equiv 1$ $(\bmod p)$. Thus $\omega$ divides $p-1$. We deduce that $\omega$ divides $\operatorname{GCD}(2 n, p-1)$. If we further impose that $p$ is the smallest prime factor of $n$, then necessarily $\omega=1$ or 2 (since a prime factor of $p-1$ cannot divide $n$ by minimality). In the first case, $3 \equiv 1(\bmod p)$ and thus $p=2$, which is excluded. In the second case, $3^{2} \equiv 1(\bmod p)$ and thus $p$ divides 8, which is also excluded. We conclude that $n=1$.