N4 (6-1, Czechoslovakia) (1) Determine all positive integers $n$, such that $2^{n}-1$ is divisible by 7;
(2) Prove that for all positive integers $n, 2^{n}+1$ is not divisible by 7.
(1) When $n=3 k$, we have
$$
2^{n}-1=8^{k}-1 \equiv 1^{k}-1=0 \quad(\bmod 7) ;
$$

When $n=3 k+1$, we have
$$
2^{n}-1=2 \cdot 8^{k}-1 \equiv 2-1=1 \quad(\bmod 7) ;
$$

When $n=3 k+2$, we have
$$
2^{n}-1=4 \cdot 8^{k}-1 \equiv 4-1=3 \quad(\bmod 7) .
$$

In summary, $7 \mid 2^n-1$ if and only if $3 \mid n$.
(2) When $n=3 k$, we have
$$
2^{n}+1 \equiv 8^{k}+1 \equiv 1^{k}+1=2 \quad(\bmod 7) ;
$$

When $n=3 k+1$, we have
$$
2^{n}+1 \equiv 2 \cdot 8^{k}+1 \equiv 2+1=3 \quad(\bmod 7) ;
$$

When $n=3 k+2$, we have
$$
2^{n}+1 \equiv 4 \cdot 8^{k}+1 \equiv 4+1=5 \quad(\bmod 7) .
$$

Therefore, $7 \nmid 2^{n}+1$.