What are the examples of “tame” minimal Kan simplicial sets having finite number of simplexes in each dimension besides simplicial point and $B(G)\approx K(G,1) $ for a finite group $G$? I believe that Alain Connes’ simplicial circle is also minimal Kan.
A tame minimal Kan complex has finite homotopy groups. The converse is also true: if homotopy groups of a minimal Kan complex are finite, then there are finitely many $n$simplices. The proof is by induction on $n$. First, $X_0 = \pi_0(X)$ is finite. If $n > 0$, there are finitely many possible choices of boundaries for $n$simplices by induction hypothesis. For every such boundary, the set of its fillers is either empty or isomorphic to $\pi_n(X,x_0)$ for some $x_0$. This completes the proof. This fact implies that, for every Kan complex with finite homotopy group, there is an equivalent tame minimal Kan complex.

$\begingroup$ Tank you Valery. I should understand better why the tame complex has finite homotopy groups. It should be some very basic knowledge which i am missing for the moment. Therefore Connes’ circle (which is $K(Z,1) $ can’t be minimal Kan? Good to know. $\endgroup$ – Nikolai Mnev Jul 7 '19 at 7:33

1$\begingroup$ @NikolaiMnev This is true for any Kan complex $X$. The $n$th homotopy group at $x_0$ could be defined as the set of homotopy classes of $n$cells of $X$ with the degenerate at $x_0$ boundary. So, there is a surjection from a subset of $X_n$ onto $\pi_n(X,x_0)$. As a side note, if $X$ is minimal, then the surjection is identity, so $\pi_n(X,x_0)$ is a subset of $X_n$. I don't know what is Connes' circle, but minimal Kan complexes of a given homotopy type are unique up to isomorphism, so if it is not isomorphic to the usual definition of $\mathbf{B}(\mathbb{Z})$, it can't be minimal Kan. $\endgroup$ – Valery Isaev Jul 7 '19 at 7:58