25. E.

Let $\angle C_{n} A_{n} B_{n}=\alpha_{n}, \angle A_{n} B_{n} C_{n}=\beta_{n}$, and $\angle A_{n} C_{n} B_{n}=\gamma_{n}$.
From the problem, we know that $A_{0}, B_{0}, A_{1}, B_{1}$ are concyclic.
Thus, $\angle A_{0} A_{1} B_{1}=\angle A_{0} B_{0} B_{1}=90^{\circ}-\alpha_{0}$.
Similarly,
$$
\angle A_{0} A_{1} C_{1}=\angle A_{0} C_{0} C_{1}=90^{\circ}-\alpha_{0}.
$$

Then $\alpha_{1}=\angle A_{0} A_{1} B_{1}+\angle A_{0} A_{1} C_{1}$
$$
=180^{\circ}-2 \alpha_{0}.
$$

Similarly, $\beta_{1}=180^{\circ}-2 \beta_{0}, \gamma_{1}=180^{\circ}-2 \gamma_{0}$.
Thus, $\alpha_{n}=180^{\circ}-2 \alpha_{n-1}$
$$
\Rightarrow \alpha_{n}=(-2)^{n}\left(\alpha_{0}-60^{\circ}\right)+60^{\circ}.
$$

Similarly,
$$
\begin{array}{l}
\beta_{n}=(-2)^{n}\left(\beta_{0}-60^{\circ}\right)+60^{\circ}, \\
\gamma_{n}=(-2)^{n}\left(\gamma_{0}-60^{\circ}\right)+60^{\circ}.
\end{array}
$$

Take $\alpha_{0}=60^{\circ}, \beta_{0}=59.999^{\circ}, \gamma_{0}=60.001^{\circ}$.
Then $\alpha_{n}=60^{\circ}$,
$$
\begin{array}{l}
\beta_{n}=-(-2)^{n} \times 0.001^{\circ}+60^{\circ} \\
\gamma_{n}=(-2)^{n} \times 0.001^{\circ}+60^{\circ}.
\end{array}
$$

When $n=15$, $\beta_{n}>90^{\circ}$.
Therefore, the smallest positive integer $n$ is 15.
25. Given that the internal angles of $\triangle A_{0} B_{0} C_{0}$ are $59.999^{\circ}$, $60^{\circ}$, and $60.001^{\circ}$. Let $A_{n}\left(n \in \mathbf{Z}_{+}\right)$ be the projection of point $A_{n-1}$ on side $B_{n-1} C_{n-1}$, $B_{n}$ and $C_{n}$ be the projections of points $B_{n-1}$ and $C_{n-1}$ on sides $C_{n-1} A_{n-1}$ and $A_{n-1} B_{n-1}$, respectively. If $\triangle A_{n} B_{n} C_{n}$ is an obtuse triangle, then the minimum value of $n$ is ).
(A) 10
(B) 11
(C) 13
(D) 14
(E) 15