1. Given an integer $k \geqslant 2$. Find all positive integers $n_{1}, n_{2}, \cdots, n_{k}$, such that:
$$
\begin{array}{l}
n_{2}\left|\left(2^{n_{1}}-1\right), n_{3}\right|\left(2^{n_{2}}-1\right), \cdots, \\
n_{k}\left|\left(2^{n_{k-1}}-1\right), n_{1}\right|\left(2^{n_{k}}-1\right) .
\end{array}
$$
1. The qualified numbers are
$$
n_{1}=n_{2}=\cdots=n_{k}=1 \text {. }
$$

For an integer $m$, let $p(m)$ denote the smallest prime factor of $m$.

Next, we prove: For any odd integers $m > 1$ and $l > 1$, if $m \mid\left(2^{l}-1\right)$, then
$$
p(l)1$, we know
$$
(l, p(m)-1)>1 \text {. }
$$

Thus, $p(l) \leqslant(l, p(m)-1) \leqslant p(m)-11$, obviously, it implies
$$
n_{i}>1(i=1,2, \cdots, k) \text {. }
$$

Therefore, $p\left(n_{1}\right)<p\left(n_{2}\right)<\cdots<p\left(n_{k}\right)<p\left(n_{1}\right)$.
Contradiction.
Thus, $n_{1}=1$, and it is easy to deduce
$$
n_{1}=n_{2}=\cdots=n_{k}=1 .
$$