jmc

Observe $A_n=a(1+10+\cdots +10^{n-1})=a\cdot \frac{10^n-1}{9}$; similarly $B_n=b\cdot \frac{10^n-1}{9}$ and $C_n=c\cdot \frac{10^{2n}-1}{9}$. The relation $C_n-B_n=A_n^2$ rewrites as$$c\cdot \frac{10^{2n}-1}{9}-b\cdot \frac{10^n-1}{9}=a^2\cdot {\left(\frac{10^n-1}{9}\right)}^2.$$Since $n>0$, $10^n>1$ and we may cancel out a factor of $\frac{10^n-1}{9}$ to obtain$$c\cdot (10^n+1)-b=a^2\cdot \frac{10^n-1}{9}.$$This is a linear equation in $10^n$. Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Now we plug in $n=0$ and $n=1$ (or some other number), we get $2c-b=0$ and $11c-b=a^2$ . Solving the equations for $c$ and $b$, we get$$c=\frac{a^2}{9}\text{and}c-b=-\frac{a^2}{9}⟹b=\frac{2a^2}{9}.$$To maximize $a+b+c=a+\frac{a^2}{3}$, we need to maximize $a$. Since $b$ and $c$ must be integers, $a$ must be a multiple of $3$. If $a=9$ then $b$ exceeds $9$. However, if $a=6$ then $b=8$ and $c=4$ for an answer of $\boxed{18}$.