$1 \cdot 33$ Find all natural numbers $n$ such that $2^{n}+1$ is divisible by 3.
(Hungarian Mathematical Olympiad, 1898)
[Solution] From $2^{n}-(-1)^{n}=3\left[2^{n-1}-2^{n-2}+\cdots+(-1)^{n-1}\right]$. For convenience, let $2^{n-1}-2^{n-2}+\cdots+(-1)^{n-1}=A$, then $A$ is an integer, so
$$
\begin{array}{c}
2^{n}-(-1)^{n}=3 A . \\
2^{n}+1=3 A+1+(-1)^{n} .
\end{array}
$$

Therefore, if and only if $n$ is odd,
$$
3 \mid 2^{n}+1 \text {. }
$$