smc

By geometric series, we have \begin{alignat}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat} By substitution, we rewrite the given equation $C_n-B_n=A_n^2$ 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 $n>0,$ it follows that $10^n>1.$ We divide both sides by $\frac{10^n-1}{9}$ and then rearrange: $$\begin{array}{rlrl}c\left(10^n+1\right)-b& =a^2\cdot \frac{10^n-1}{9}& & \\ 9c\left(10^n+1\right)-9b& =a^2\left(10^n-1\right)& & \\ \left(9c-a^2\right)10^n& =9b-9c-a^2.& & (★)\end{array}$$ Let $y=10^n.$ Note that $(★)$ is a linear equation with $y,$ and $y$ is a one-to-one function of $n.$ Since $(★)$ has at least two solutions of $n,$ it has at least two solutions of $y.$ We conclude that $(★)$ must be an identity, so we get the following system of equations: $$\begin{array}{rl}9c-a^2& =0,\\ 9b-9c-a^2& =0.\end{array}$$ The first equation implies that $c=\frac{a^2}{9}.$ Substituting this into the second equation gives $b=\frac{2a^2}{9}.$ To maximize $a+b+c=a+\frac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a+b+c$ is $6+8+4=\boxed{18}.$