We claim $c_{\min} = 2$.
Taking $a_j = \frac{1}{j}$ for $j = 1, 2, \dots, n$, we have
$$\sum_{k=1}^{n} \frac{k}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k}} = \sum_{k=1}^{n} \frac{k}{1 + 2 + \dots + k} = 2 \sum_{k=1}^{n} \frac{1}{k+1}, \text{ while } c \sum_{k=1}^{n} a_k =$$

We will now prove that $c = 2$ is suitable. From the Cauchy-Schwartz inequality,
$$\frac{k^2(k+1)^2}{4} = \left(\sum_{j=1}^{k} j\right)^2 \le \left(\sum_{j=1}^{k} j^2 a_j\right) \left(\sum_{j=1}^{k} \frac{1}{a_j}\right), \text{ hence} \\ \frac{k}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k}} \le \frac{4}{k(k+1)^2} \sum_{j=1}^{k} j^2 a_j.$$
Therefore
$$\begin{align*} \sum_{k=1}^{n} \frac{k}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k}} &\le \sum_{k=1}^{n} \left( \frac{4}{k(k+1)^2} \sum_{j=1}^{k} j^2 a_j \right) = \\ &= \sum_{j=1}^{n} \left( j^2 a_j \sum_{k=j}^{n} \frac{4}{k(k+1)^2} \right) = 2 \sum_{j=1}^{n} \left( j^2 a_j \sum_{k=j}^{n} \frac{2k}{k^2(k+1)^2} \right) < \\ &< 2 \sum_{j=1}^{n} \left( j^2 a_j \sum_{k=j}^{n} \frac{2k+1}{k^2(k+1)^2} \right). \text{ But} \\ \sum_{k=j}^{n} \frac{2k+1}{k^2(k+1)^2} &= \sum_{k=j}^{n} \left( \frac{1}{k^2} - \frac{1}{(k+1)^2} \right) = \frac{1}{j^2} - \frac{1}{(n+1)^2} < \frac{1}{j^2}, \\ \text{hence } \sum_{k=1}^{n} \frac{k}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k}} &< 2 \sum_{k=1}^{n} a_k. \end{align*}$$