2022 AMC 12A Problem 23

$L_n$ is always divisible by $k_n,$ so $k_n\lt L_n$ exactly when some prime $p$ divides both $L_n$ and the numerator $N=\sum_{k=1}^n \tfrac{L_n}{k}$ (i.e. a prime cancels). For a prime $p$ with maximal power $p^a\le n,$ only the terms with $v_p(k)=a$ keep $p$ out of $\frac{L_n}{k};$ all others are divisible by $p.$ So $p$ cancels iff $\sum_{v_p(k)=a}\tfrac{L_n}{k}\equiv0\pmod p.$ Applying this test recursively with $H_n=H_{n-1}+\frac{1}{n}$ gives $k_n=L_n$ for $1\le n\le5,$ $k_n=\frac{L_n}{3}$ for $6\le n\le8,$ and $k_n=L_n$ again for $9\le n\le17.$ Finally, the ratios $\frac{L_n}{k_n}$ for $n=18,19,20,21,22$ are $3,3,15,45,45,$ respectively. Thus cancellation occurs precisely for $n=6,7,8,18,19,20,21,22,$ which is $8$ values. Thus, the correct answer is D.