We claim that the values of $ n$ such that every $ (n, k)-\text{cloud}$ has two mutually exterior tangent circumferences of the same color are the naturals that satisfy $ n\geq 2^k+1$. Let us identify a $ (n, k)-\text{cloud}$ with a $ n\times n$ matrix with entries in $ \{1,\cdots,k\}$ (without the diagonal), being the element $ a_{i,j}$ the color of the circle with diameter $ P_iP_j$ (considering WLOG that the set of colors is $ \{1,\cdots,k\}$). Then a $ (n, k)-\text{cloud}$ doesn't have two mutually exterior tangent circumferences of the same color iff $ \forall \, i$, $ \{a_{i,1},\cdots,a_{i,i-1}\}$ e $ \{a_{i,i+1},\cdots,a_{i,n}\}$ are disjoint sets. Observation: For us to prove that it is possible to have a $ (n, k)-\text{cloud}$ without two mutually exterior tangent circumferences of the same color for $ n\leq 2^k$ it is enough to construct such an example for $ n=2^k$ (because from that example we immediately get such a cloud for smaller values of $ n$). We will prove this by induction on $ k$. Base case - $ k=1$ $ \begin{bmatrix} \times & 1 \\ 1 & \times \end{bmatrix}$ Induction step: Suppose that we have a matrix $ M$ for $ k$, and let $ N=\begin{bmatrix} k+1 & k+1 & \cdots & k+1 \\ k+1 & k+1 & \cdots & k+1 \\ \vdots & \vdots & \ddots & \vdots \\ k+1 & k+1 & \cdots & k+1\end{bmatrix}$ be a $ 2^k\times 2^k$ matrix with all entries being $ k+1$. Then $ \begin{bmatrix} M & N \\ N & M \end{bmatrix}$ is a matrix for $ k+1$, since it is symmetric and if $ i\leq 2^k$, then $ \{a_{i1},\cdots,a_{i,i-1}\}$ and $ \{a_{i,i+1},\cdots,a_{i,2^k}\}$ are disjoint by the induction hypothesis, and $ k+1\not \in \{a_{i,1},\cdots,a_{i,i-1}\}$ (the case $ i>2^k$ is analogous). We will prove now that for any such matrix $ n\times n$ with $ n\geq 2^k+1$ there exists $ i$ such that $ \{a_{i,1},\cdots,a_{i,i-1}\}$ and $ \{a_{i,i+1},\cdots,a_{i,n}\}$ are not disjoint. Because the matrix is symmetrical, that is equivalent to $ \{a_{i,1},\cdots,a_{i,i-1}\}$ e $ \{a_{i+1,i},\cdots,a_{n,i}\}$ having a common element. Consider the following sets: $ \{a_{2,1}\}$ $ \{a_{3,1},a_{3,2}\}$ $ \cdots$ $ \{a_{n,1},\cdots,a_{n,n-1}\}$ Since there are $ n-1>2^k-1$ sets and $ \{1,\cdots,k\}$ only has $ 2^k-1$ non-empty subsets, by the pigeonhole principle we have that two of those sets are equal. Let they be $ \{a_{i1},\cdots,a_{i,i-1}\}$ e $ \{a_{j1},\cdots,a_{j,j-1}\}$, $ j>i$. Then $ a_{j,i}$ is in both sets (because they are equal), so $ \{a_{i1},\cdots,a_{i,i-1}\}$ e $ \{a_{i+1,i},\cdots,a_{j,i},\cdots,a_{n,i}\}$ are not disjoint, as desired.