Note: I originally misread the constraint as $\sum_{i=1}^n \frac{a_i}{b_i} \leq \sum_{i=1}^n \frac{c_i}{d_i}$, while it actually is just one of the $n$ constraints. But it turns out that the conclusion $\frac {\sum_i a_i}{\sum_i b_i} \leq \frac {\sum_i c_i}{\sum_i d_i}$ again fails, and for pretty much the same reason. So here goes the modified solution...
One way of looking at the problem is as follows. Suppose $x_i = a_i/b_i$ and $y_j = c_j/d_j$. Then the problem stipulates that $\sum_{i=1}^k x_i \leq \sum_{j=1}^k y_j$ (mistake corrected). Now, we are asked whether it is necessarily the case that $ \frac{\sum_i b_i x_i}{\sum_i b_i} \stackrel{?}{\leq} \frac{\sum_j d_j y_j}{\sum_j d_j}. $
What does each side of the conjectured inequality mean? The LHS is the "weighted average" of the quantities $x_i$, according to the weights $b_i$; similarly for the RHS. It is easy to see the following proposition.
Suppose $x_i > 0$ are a set of numbers with minimum value $m$ and maximum value $M$. Then we have $m \leq \frac{\sum_i b_i x_i}{\sum_i b_i} \leq M$ assuming $b_i \geq 0$ and $\sum_i b_i > 0$. Moreover, for any $\mu \in [m, M]$, there exists some choice of $b_i \geq 0$ such that $\sum_i b_i = 1 > 0$ and $\frac{\sum_i b_i x_i}{\sum_i b_i} = \mu$.
Added condition: Suppose $m < \mu < M$ (that is, $m < M$ and these extreme points points are no longer allowed). Then there exists some choice of the weights $b_i > 0$ such that $\sum_{i} b_i = 1 > 0$ and $\frac{\sum_i b_i x_i}{\sum_i b_i} = \mu$. The point here is that none of the weights $b_i$ are allowed to be zero.
The proof of this proposition is simple, and I encourage you to work it out.
Now, we'll find a counterexample to the given problem. Assume that $x_i > 0$ is such that $\min_i x_i < \max_i x_i$ (that is, not all $x_i$s are equal). Let $y_i := x_i$ for all $i$, so that $\sum_{i=1}^k x_i \leq \sum_{j=1}^k y_j$ trivially holds for all $k$. Show that there exists some choice of weights $b_i > 0$ and $d_j > 0$ such that $\sum_{i} b_i = \sum_j d_j = 1$ and $ \frac{\sum_i b_i x_i}{\sum_i b_i} > \frac{\sum_j d_j y_j}{\sum_j d_j} . $