Solution:
The function $K(n)$ is increasing for $n \geq 3$, since $\sum_{i=1}^{n} \frac{1}{x_{i}}=1$ implies that
$$\sum_{i=1}^{n-1} \frac{1}{x_{i}}+\frac{1}{x_{n}+1}+\frac{1}{x_{n}(x_{n}+1)}=1$$
Thus it is enough to find $t \leq L(b)$ such that $K(t+2)+2 L(b) \geq d(b)$, where $d(b)$ is the number of the distinct positive integers that divide $b$. Let $t$ be the minimal positive integer, for which the equation $\sum_{i=1}^{t} \frac{1}{x_{i}}=1-\frac{1}{b}$ has a solution. Then $t \leq L(b)$. Fix now $t$, $b$, $x_{1}, \ldots, x_{t}$.

Note that the numbers of the solution of the equation $\frac{1}{y_{1}}+\frac{1}{y_{2}}=\frac{1}{b}$ such that $b$ divides $y_{2}$ and $y_{1} \leq y_{2}$, is equal to $d(b)$. Indeed, if $\frac{1}{b}=\frac{1}{y_{1}}+\frac{1}{k b}$ for some $k \geq 2$, then $y_{1}=b+\frac{b}{k-1}$. So $k-1$ divides $b$ and there are exactly $d(b)$ possibilities for $k$.

Hence $K(t+2)$ is not less than $d(b)$ minus the number of the cases, when $y_{i}=x_{j}$. This cases are at most $2 L(b)$ which implies the desired inequality.