222. Find all natural $n$ for which the sum $2^{n}+1$ is divisible by 3.
$\triangle$ Obviously, all odd $n$ will work, since in this case the sum $2^{n}+1$ is divisible by $2+1$, i.e., by 3. And what if $n$ is even?

Let $n$ be even: $n=2 k(k \in N)$. Transform the given sum:

$$
2^{n}+1=2^{2 k}+1=\left(2^{2 k}-1\right)+(1+1)=\left(4^{k}-1\right)+2
$$

Since $4^{k}-1$ is divisible by 3, and 2 is not, the entire sum is not divisible by 3.

Answer: for all odd $n$.