By the Cauchy-Schwarz inequality, we have $$(k_1 + k_2 + \cdots + k_n)\Big(\frac{1}{k_1} + \frac{1}{k_2} + \cdots + \frac{1}{k_n}\Big) \ge n^2$$ Therefore we must have $(5n-4) \ge n^2$, which leads to $n\le 4$.

When $n=1$, we must have $k_1=1$.

When $n=2$, then $$ \left\{ \begin{array}{rcl} k_1 + k_2 &=& 6\\ \frac{1}{k_1} + \frac{1}{k_2}&=&1 \end{array} \right. $$ This system does not have positive integer solution.

When $n=3$, then $$ \left\{ \begin{array}{rcl} k_1 + k_2 +k_3&=& 11\\ \frac{1}{k_1} + \frac{1}{k_2}+\frac{1}{k_3}&=&1 \end{array} \right. $$ Without loss of generality, let's assume $k_1\le k_2\le k_3$. Then \begin{align} k_1 + k_2+k_3=11 &\implies 1\le k_1 \le 3\\ \frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=1&\implies 2 \le k_1 \le 3 \end{align} Therefore $(k_1, k_2, k_3)$ can only be $(2,2,7)$, $(2, 3, 6)$, $(2,4,5)$, $(3,3,5)$, and $(3, 4, 4)$. Among these candidates, only $(2, 3, 6)$ works.

When $n=4$, then $$ \left\{ \begin{array}{rcl} k_1 + k_2 +k_3+k_4&=& 16\\ \frac{1}{k_1} + \frac{1}{k_2}+\frac{1}{k_3}+\frac{1}{k_4}&=&1 \end{array} \right. $$ In this case, the first inequality takes equality, therefore we must have $k_1=k_2=k_3=k_4$. It is easy to see that $(4, 4, 4, 4)$ works.

Hence, in summary, all the solutions are - $n=1$ and $k_1=1$ - $n=3$ and $(k_1, k_2, k_3)$ are all the permutations of $(2, 3, 6)$ - $n=4$ and $k_1=k_2=k_3=k_4=4$