Solution:

The answer is: $2 \leq k \leq n \leq 3$ or $3 \leq k \leq n$.
Clearly $k \leq n$ holds by the statement of the problem, and also $k \geq 2$, because for $k=1$ all points would have the same color, while the number $n$ of colors is supposed to be $\geq 2$. We number the circles and the colors from 1 to $n$ and denote by $F(i, j)$ the set of colors of the intersection points of circles $i$ and $j$. $F(i, j)$ contains one or two elements.

Let $k=2$. For $n=2$, $F(1,2)=\{1,2\}$ is an admissible coloring. For $n=3$, $F(1,2)=\{3\}$, $F(2,3)=\{1\}$, $F(3,1)=\{2\}$ is an example of an admissible coloring. Now let $n \geq 4$. To each of the $n$ circles we assign the set $\{i, j\}$ of the two colors occurring on it. Each of these sets consists of two elements, and each of the $n$ colors must occur in at least two sets, since at every colored point two circles intersect. Hence each color occurs in exactly two sets. For circle 1 with the set $\{i, j\}$ there are therefore at most two other circles in whose color sets $i$ or $j$ occur. Since $n \geq 4$, we can always find a circle 2 with the set $\{k, l\}$ and $\{k, l\} \cap \{i, j\}=\{ \}$. The intersection points of circles 1 and 2 then cannot be colored admissibly – contradiction!

We now prove by complete induction a somewhat stronger statement than required: For $n \geq k \geq 3$ there always exists an admissible coloring in which, on circle $i$, the color $i$ occurs, for all $i=1, \ldots, n$. As a base case, for $k=n=3$ we give the example $F(1,2)=\{1,2\}$, $F(1,3)=\{1,3\}$, $F(2,3)=\{2,3\}$, and for $k=3, n>3$ the following example of an admissible coloring with the additional condition:

$F(1,2)=\{1,2\}$, $F(i, i+1)=\{i\}$ for $1<i<n-1$, $F(n-1, n)=\{n-2, n-1\}$ and $F(i, j)=\{n\}$ for the remaining pairs $(i, j)$ with $1 \leq i<j \leq n$.

Now we assume that the stronger statement holds for some $k \geq 3$, and choose $n \geq k+1$. Since $n-1 \geq k \geq 3$, there is an admissible coloring with the additional condition for the circles resp. colors $1,2,..., n-1$. Now we color the intersection points of circle $n$: for each $i=1, \ldots, n-1$, one intersection point of circles $i$ and $n$ receives the color $n$. Thus on each of the circles $i$ with $i=1, \ldots, n-1$ exactly $k+1$ colors occur; among them $i$ and $n$. For $i=1, \ldots, k$, the second intersection point of circles $i$ and $n$ receives the color $i$, so that now exactly $k+1$ colors also lie on circle $n$, namely $1$ through $k$ and $n$. All remaining new intersection points receive the color $n$, so that no further colors are added on any circle. This coloring satisfies all conditions.