Answer: E

Solution:

Take $k=2j.$ The smallest perfect square with $k+1$ digits is $10^{k}=(10^{j})^2,$ and after Silvia erases, the numbers shown are $\left\lfloor n^2/10^{k}\right\rfloor$ for $n=10^{j}, 10^{j}+1,\ldots$

Put $M=10^k.$ A jump of at least $2$ from $n$ to $n+1$ requires $(n+1)^2-n^2=2n+1\gt M,$ so write $n=M/2+m$ with $m\ge0.$ The case $m=0$ gives a jump of only $1,$ so the first larger jump has $m\ge1.$ Let $A=\left\lfloor n^2/M\right\rfloor$ and $B=\left\lfloor(n+1)^2/M\right\rfloor.$ Because $M$ is divisible by $4,$ $$\begin{aligned} A&=\dfrac M4+m+\left\lfloor\dfrac{m^2}{M}\right\rfloor,\\ B&=\dfrac M4+m+1\\ &\quad{}+\left\lfloor\dfrac{(m+1)^2}{M}\right\rfloor. \end{aligned}$$

Therefore the first jump of at least $2$ occurs at the first $m$ for which $m^2\lt M\le(m+1)^2.$ Since $\sqrt M=10^j,$ this is $m=10^j-1.$ The last displayed value before the gap is $M/4+10^j-1,$ so the smallest missing integer is $$f(2j)=\dfrac{10^{2j}}4+10^j.$$

Summing over $j=1,\ldots,1008,$ $$\begin{gathered} \sum_{j=1}^{1008}f(2j)\\ =25\sum_{j=0}^{1007}10^{2j}\\ {}+10\sum_{j=0}^{1007}10^{j}\\ =\underbrace{2525\cdots25}_{2016\text{ digits}}\\ {}+\underbrace{111\cdots10}_{1009\text{ digits}}. \end{gathered}$$ There are no carries, so the digit sum is $1008\cdot(2+5)$ $+1008\cdot 1=1008\cdot 8=8064.$

Thus, the correct answer is E.