25. Let $n \in \mathbf{Z}_{+}$,
```
\begin{array}{l}
a 、 b 、 c \in\{x \mid x \in \underbrace{\mathbf{Z}, \text { and }}_{n} x \in[\underline{1,9]\}}, \\
A_{n}=\underbrace{\overline{a a \cdots a}}_{n \uparrow}, B_{n}=\underbrace{b b \cdots b}_{2 n \uparrow}, C_{n}=\underbrace{c c \cdots}_{2 n \cdots c} .
\end{array}
```

Then the maximum value of $a+b+c$ is ( ) when there exist at least two $n$ such that $C_{n}-B_{n}=A_{n}^{2}$.
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
25. D.

Notice,
$$
\begin{array}{l}
A_{n}=\underbrace{\overline{a a \cdots a}}_{n \uparrow}=a \cdot \frac{10^{n}-1}{9}, \\
B_{n}=\underbrace{\overline{b b \cdots b}}_{n \uparrow}=b \cdot \frac{10^{n}-1}{9}, \\
C_{n}=\underbrace{\overline{c c \cdots c}}_{2 n \uparrow}=c \cdot \frac{10^{2 n}-1}{9} .
\end{array}
$$

From $C_{n}-B_{n}=A_{n}^{2}$, we know
$$
\begin{array}{l}
c \cdot \frac{10^{2 n}-1}{9}-b \cdot \frac{10^{n}-1}{9} \\
\quad=\left(a \cdot \frac{10^{n}-1}{9}\right)^{2}=a^{2}\left(\frac{10^{n}-1}{9}\right)^{2} \\
\Rightarrow 9\left(c\left(10^{n}+1\right)-b\right)\left(10^{n}-1\right) \\
\quad=a^{2}\left(10^{n}-1\right)^{2} .
\end{array}
$$

Since there are at least two $n$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
10^{n}-1 \neq 0, \\
9 c\left(10^{n}+1\right)-9 b=a^{2}\left(10^{n}-1\right)
\end{array}\right. \\
\Rightarrow\left(9 c-a^{2}\right) 10^{n}+9 c-9 b+a^{2}=0 \\
\Rightarrow\left\{\begin{array}{l}
a^{2}=9 c, \\
a^{2}+9 c-9 b=0
\end{array}\right. \\
\Rightarrow 3 \mid a \text {. } \\
\end{array}
$$

Upon inspection, when $a=6$, $b=8$, $c=4$, at this time, $a+b+c=18$ is the maximum value.