Definition: if $a$ is a back fragment of $b$, then $b$ is an initial fragment of $a$. Lemma: $x$ is a positive integer and $x_1,...,x_n$ are back fragments of $x$. Also none of them is an initial fragment of another. Then $\frac{1}{x}\geq \frac{1}{x_1}+...+\frac{1}{x_n}$. Proof: $\frac{1}{x}>\frac{1}{10x}+\frac{1}{10x+1}+...+\frac{1}{10x+9}$. Then all of $\frac{1}{x_i}$ are either these 10 terms or are back fragments. But if say $x_1=10x+1$, then the rest are not a back fragment of $10x+1$. So we repeat the first inequality as many times until we are done. Example: $10,11,121,199$ are back fragments of 1. Then \[\frac{1}{1}>\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{10}+\frac{1}{11}+\frac{1}{120}+\frac{1}{121}+...+\frac{1}{129}+\frac{1}{13}+...+\frac{1}{18}+\frac{1}{190}+\frac{1}{191}+...+\frac{1}{199}\] So now back to the question, all numbers are back fragments of one of $1,2,...,9$. Thus by the lemma $\sum{\frac{1}{x_i}}\leq\sum_{i=1}^{9}{\frac1i}<3$. Hope it can be understood.