smc

We are given that $$\sum _{i=1}^n\frac{1}{i}=\frac{1}{L_n}\sum _{i=1}^n\frac{L_n}{i}=\frac{h_n}{k_n}.$$ Since $k_n<L_n,$ we need $\gcd \left({\sum }_{i=1}^n\frac{L_n}{i},L_n\right)>1.$ For all primes $p$ such that $p\le n,$ let $v_p(L_n)=e\ge 1$ be the largest power of $p$ that is a factor of $L_n.$ It is clear that $L_n\equiv 0(\mathrm{mod}p),$ so we test whether ${\sum }_{i=1}^n\frac{L_n}{i}\equiv 0(\mathrm{mod}p).$ Note that $$\sum _{i=1}^n\frac{L_n}{i}\equiv \sum _{i=1}^{\lfloor \frac{n}{p^e}\rfloor }\frac{L_n}{i{p}^e}(\mathrm{mod}{p}^e)\equiv \sum _{i=1}^{\lfloor \frac{n}{p^e}\rfloor }\frac{L_n}{i{p}^e}(\mathrm{mod}p).$$ We construct the following table for $v_p(L_n)=e:$ \begin{array}{c|c|l|c} \textbf{Case of }\boldsymbol{(p,e)} & \textbf{Interval of }\boldsymbol{n} & \hspace{22.75mm}\textbf{Sum} & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline & & & \\ [-2ex] (2,1) & [2,3] & L_n/2 & \\ (2,2) & [4,7] & L_n/4 & \\ (2,3) & [8,15] & L_n/8 & \\ (2,4) & [16,22] & L_n/16 & \\ [0.5ex] \hline & & & \\ [-2ex] (3,1) & [3,5] & L_n/3 & \\ & [6,8] & L_n/3 + L_n/6 & \checkmark \\ (3,2) & [9,17] & L_n/9 & \\ & [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (5,1) & [5,9] & L_n/5 & \\ & [10,14] & L_n/5 + L_n/10 & \\ & [15,19] & L_n/5 + L_n/10 + L_n/15 & \\ & [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (7,1) & [7,13] & L_n/7 & \\ & [14,20] & L_n/7 + L_n/14 & \\ & [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex] \hline & & & \\ [-2ex] (11,1) & [11,21] & L_n/11 & \\ & \{22\} & L_n/11 + L_n/22 & \\ [0.5ex] \hline & & & \\ [-2ex] (13,1) & [13,22] & L_n/13 & \\ [0.5ex] \hline & & & \\ [-2ex] (17,1) & [17,22] & L_n/17 & \\ [0.5ex] \hline & & & \\ [-2ex] (19,1) & [19,22] & L_n/19 & \\ [0.5ex] \end{array} Note that: 1. If the Sum column has only one term, then it is never congruent to $0$ modulo $p.$ 2. If $p$ and $q$ are positive integers such that $p\ge q,$ then $L_p$ is a multiple of $L_q.$ Therefore, for a specific case, if the sum is congruent to $0$ modulo $p$ for the smallest element in the interval of $n,$ then it is also congruent to $0$ modulo $p$ for all other elements in the interval of $n.$ Together, there are $\boxed{8}$ such integers $n,$ namely $$6,7,8,18,19,20,21,22.$$