Let $\{y_i\}_{i=1}^n$ be a sequence of positive real numbers such that $\sum_{i=1}^ny_i \ge Cn$ with $C > 0 $, for every $n$. Let $\{x_i\}_{i=1}^n$ be a binary sequence $x_i \in \{0,1\}$ such that $\sum_{i=1}^nx_i \ge C'n$ with $C' > 0$, for every $n$. Is it the case that $$\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^ny_i}> C''$$ with $C'' > 0$, for every $n$? If not, for $n$ sufficiently large? My intuition says that this should be true, but I am struggling to come up with a formal argument.
EDIT: as the counterexample below shows, $y_i >0$ is too weak of an assumption, let me suppose instead that $y_i > c$ for some $c > 0$.
No. Let $(y_n)_n$ be defined by $$ y_n = \begin{cases}1 & \text{ if } n \text{ odd}\\\frac{1}{2^n} & \text{ otherwise.}\end{cases} $$ and $(x_n)_n$ by $$ x_n = \begin{cases}0 & \text{ if } n > 1 \text{ odd}\\1 & \text{ otherwise.}\end{cases} $$
Then both assumptions are satisfied (with $C=C'=1/2$), yet $$\forall n\geq 1,\qquad\frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n y_i} = \frac{1}{\sum_{i=1}^n y_i} \xrightarrow[n\to\infty]{} 0\,.$$
To deal with the edited question: the ratio above is invariant by scaling (of $(y_n)_n$), so replacing $y_n$ by $y'_n = 2^ny_n$ (now bounded away from $0$) leads to the same conclusion.
Note: I had to choose $x_1=1$ as a technicality for the assumption on $(x_n)_n$ to hold even for $n=1$.