For an $n$-tuple of integers, define a transformation to be:

$$(a_1,a_2,\cdots,a_{n-1},a_n)\rightarrow (a_1+a_2, a_2+a_3, \cdots, a_{n-1}+a_n, a_n+a_1)$$

Find all ordered pairs of integers $(n,k)$ with $n,k\geq 2$, such that for any $n$-tuple of integers $(a_1,a_2,\cdots,a_{n-1},a_n)$, after a finite number of transformations, every element in the of the $n$-tuple is a multiple of $k$.
1. **Define the transformation and the problem:**
   Given an $n$-tuple of integers $(a_1, a_2, \cdots, a_{n-1}, a_n)$, the transformation is defined as:
   \[
   (a_1, a_2, \cdots, a_{n-1}, a_n) \rightarrow (a_1 + a_2, a_2 + a_3, \cdots, a_{n-1} + a_n, a_n + a_1)
   \]
   We need to find all ordered pairs of integers $(n, k)$ with $n, k \geq 2$ such that after a finite number of transformations, every element in the $n$-tuple is a multiple of $k$.

2. **Claim that $k \geq 3$ is not possible:**
   To show this, we need to analyze the transformation and its effect on the $n$-tuple. Denote the $k$-th transformation as $\delta(n)$.

3. **Use Fibonacci numbers to express the transformation:**
   Let $f_n$ be the $n$-th Fibonacci number. The $k$-th transformation can be expressed as:
   \[
   \delta(n) = \{f_n a_1 + f_{n-1} \sum_{j=2}^n a_j, \cdots, f_n a_n + f_{n-1} \sum_{j=1}^{n-1} a_j\}
   \]
   This implies that:
   \[
   n \mid f_n (a_n - a_{n-1}) - f_{n-1} (a_{n-1} - a_{n-2})
   \]
   This expression is invariant under the transformation, meaning it does not change (or its change can be controlled).

4. **Show that for $n = 2$, any $k$ works:**
   Consider the case when $n = 2$. The transformation becomes:
   \[
   (a_1, a_2) \rightarrow (a_1 + a_2, a_1 + a_2)
   \]
   After one transformation, both elements of the tuple are equal. Repeating the transformation, we get:
   \[
   (a_1 + a_2, a_1 + a_2) \rightarrow (2(a_1 + a_2), 2(a_1 + a_2)) \rightarrow \cdots \rightarrow (k(a_1 + a_2), k(a_1 + a_2))
   \]
   Hence, for $n = 2$, any $k$ works.

5. **Conclusion:**
   The only ordered pairs $(n, k)$ that satisfy the condition are $(2, k)$ for any $k \geq 2$.

The final answer is $\boxed{(n, k) = (2, k)}$.