4. Define a transformation of an $n$-element integer array as
$$
\begin{array}{l}
\left(a_{1}, a_{2}, \cdots, a_{n-1}, a_{n}\right) \\
\rightarrow\left(a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n-1}+a_{n}, a_{n}+a_{1}\right) .
\end{array}
$$

Find all pairs of positive integers $(n, k) (n, k \geqslant 2)$, such that for any $n$-element integer array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, after a finite number of transformations, every number in the resulting array is a multiple of $k$.
（Zhang Xinzhe, problem contributor）
4. $(n, k)=\left(2^{p}, 2^{q}\right)\left(p 、 q \in \mathbf{Z}_{+}\right)$.

First, we prove a lemma.
Lemma: Let the integer array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ after $t$ transformations be $\left(a_{1}^{(t)}, a_{2}^{(t)}, \cdots, a_{n}^{(t)}\right)$. Then $a_{i}^{(t)}=\sum_{j=0}^{t} a_{i+j} \mathrm{C}_{t}^{j}(i=1,2, \cdots, n)$.
Proof: By mathematical induction.
When $t=1$, the conclusion is obviously true.
Assume $a_{i}^{(t)}=\sum_{j=0}^{t} a_{i+j} \mathrm{C}_{t}^{j}(i=1,2, \cdots, n)$.
Then $a_{i}^{(t+1)}=a_{i}^{(t)}+a_{i+1}^{(t)}$
$=\sum_{j=0}^{t} a_{i+j} \mathrm{C}_{t}^{j}+\sum_{j=0}^{t} a_{i+1+j} \mathrm{C}_{t}^{j}$
$=a_{i} \mathrm{C}_{t}^{0}+\sum_{j=1}^{t} a_{i+j}\left(\mathrm{C}_{t}^{j}+\mathrm{C}_{t}^{j-1}\right)+a_{i+1+t} \mathrm{C}_{t}^{t}$
$=\sum_{j=0}^{t+1} a_{i+j} \mathrm{C}_{t+1}^{j}$.
The lemma is proved.
Next, we prove that $n$ and $k$ are both powers of 2.
Notice that, after each transformation, the sum of the $n$ numbers is twice the sum of the original $n$ numbers.

Let $a_{1}=1, a_{2}=a_{3}=\cdots=a_{n}=0$. By the problem statement, after a finite number of transformations (let's say $m$ times), each number is a multiple of $k$. From the above, after $m$ transformations, the sum of the $n$ numbers is $2^{m}$. Therefore, $k \mid 2^{m}$, which means $k$ is a power of 2. Thus, after $m$ transformations, each number is a multiple of 2. Consequently, each number after each subsequent transformation is also a multiple of 2.
Choose $2^{s}>m\left(s \in \mathbf{Z}_{+}\right)$.
Notice that,
$\mathrm{C}_{2^{s}}^{i}=\frac{2^{s}}{i} \mathrm{C}_{2^{s}-1}^{i-1}\left(1 \leqslant i \leqslant 2^{s}-1\right)$ is even.
Then after $2^{s}$ transformations,
$$
a_{1}^{\left(2^{s}\right)} \equiv a_{1}+a_{1+2^{s}} \equiv 0(\bmod 2) \text {. }
$$

Thus, $a_{1+2^{s}}=1=a_{1}$.
Therefore, $n \mid 2^{s}$. Hence, $n$ is a power of 2.
Next, we show that when
$$
(n, k)=\left(2^{p}, 2^{q}\right)\left(p 、 q \in \mathbf{Z}_{+}\right)
$$

any $n$-element integer array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ can be transformed into an array where each number is a multiple of $k$ after a finite number of transformations.

In fact, combining the lemma and conclusion (1), for the array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ after $n=2^{p}$ transformations, we have
$$
a_{i}^{(n)} \equiv a_{i}+a_{i+n} \equiv 0(\bmod 2)(i=1,2, \cdots, n) .
$$

Then, for the array $\left(\frac{1}{2} a_{1}^{(n)}, \frac{1}{2} a_{2}^{(n)}, \cdots, \frac{1}{2} a_{n}^{(n)}\right)$ after $n=2^{p}$ transformations, each number is also even, i.e.,
$$
a_{i}^{(2 n)} \equiv 0(\bmod 4)(i=1,2, \cdots, n) \text {. }
$$

By the principle of induction, we have
$$
a_{i}^{(g n)} \equiv 0\left(\bmod 2^{q}\right)(i=1,2, \cdots, n),
$$

which means each number is a multiple of $k=2^{q}$.
In conclusion, the statement is true.