Does there always exist real numbers $w_{1},\dots, w_{n} > 0$ of such a kind? Let $a_{1}, \dots, a_{n}$ be real numbers not all zero; let $b_{1},\dots, b_{n}$ be real numbers; let $\sum_{1}^{n}b_{i} \neq 0$. Then does there exist real numbers $w_{1},\dots, w_{n} > 0$ such that 
$$
\frac{\sum_{1}^{n}w_{i}a_{i}}{\sum_{1}^{n}w_{i}b_{i}} > \frac{\sum_{1}^{n}a_{i}}{\sum_{1}^{n}b_{i}}?
$$
Some function theory results seem prominent. But it seems that perhaps such a result is not in my current set of working knowledge. 
 A: Let $U = \{\mathrm{w} \in \mathbb{R}_+^n : \langle \mathrm{w}, \mathrm{b} \rangle \neq 0 \}$ and define $f : U \to \mathbb{R}$ by
$$ f(\mathrm{w}) = \frac{\langle \mathrm{w}, \mathrm{a} \rangle}{\langle \mathrm{w}, \mathrm{b} \rangle}. $$
We know that $U$ is open and $\mathbf{1} = (1,\cdots,1) \in U$. Let as assume that $\mathrm{a}$ and $\mathrm{b}$ are not parallel. Then
$$ \left. \frac{\partial f}{\partial w_i} \right|_{\mathrm{w}=\mathbf{1}} =  \frac{\langle \mathbf{1}, \mathrm{b} \rangle a_i - \langle \mathbf{1}, \mathrm{a} \rangle b_i}{\langle \mathbf{1}, \mathrm{b} \rangle^2} $$
The assumption tells that not all $\partial f_i / \partial w_i$ vanish. So $\nabla f(\mathbf{1})$ is non-zero. Therefore
$$ f( \mathbf{1} + \delta \nabla f(\mathbf{1})) = f(\mathbf{1}) + \| \nabla f (\mathbf{1}) \|^2 \delta + \mathcal{O}(\delta^2) \quad \text{as } \delta \downarrow 0 $$
and by taking sufficiently small $\delta > 0$ we can find $\mathrm{w} \in U$ with $f(\mathrm{w}) > f(\mathbf{1})$.
A: $\DeclareMathOperator{\spn}{span}\DeclareMathOperator{\img}{Im}$There is really some linear algebra hidden there. Note that $\sum_{i=1}^{n}w_{i}a_{i}$ is a linear functional on $\mathbb{R}^n$ with respect to $w$. That is, $\alpha: \mathbb{R}^n \to \mathbb{R} : (w_1, \cdots, w_n) \mapsto \sum_{i=1}^{n}w_{i}a_{i}$ is linear. Likewise, let $\beta: \mathbb{R}^n \to \mathbb{R} : (w_1, \cdots, w_n) \mapsto \sum_{i=1}^{n}w_{i}b_{i}$. Also let $w = (w_1, \cdots, w_n)$ and $f : D \to \mathbb{R} : w \mapsto \frac{\alpha(w)}{\beta(w)}$. Thus your question in this linear algebra framework is

Let $\alpha, \beta \in \mathcal{L}(\mathbb{R}^n, \mathbb{R})$ such that $\alpha \neq 0$ and $\beta(\vec{1}) \neq 0 $. Then there exists $w \in \mathbb{R}^n$ such that $$f(w) > f(\vec{1})? \tag{1}$$

The domain of $f$ (undetermined so far) is simply all $w \in \mathbb{R}^n$ such that $\beta(w) \neq 0$, that is, $D = \mathbb{R}^n \setminus \ker(\beta)$. By hypothesis we have $\beta(\vec{1}) \neq 0$, therefore $D = \mathbb{R}^n \setminus \spn(\vec{1})^{\perp}$.
Furthermore, an additional condition is required. It is necessary that $\alpha$ and $\beta$ are linearly independent, meaning that must not exist a $\lambda \in \mathbb{R}$ such that $\alpha = \lambda \beta$, because otherwise $f$ would be constant and $(1)$ would never be satisfied. This is equivalent to require $\ker(\alpha) \neq \ker(\beta) = \spn(\vec{1})^{\perp}$.
Now, since $\img(\alpha) = \img(\beta) = \mathbb{R}$ and for every sequence $(x_n)_{n \in \mathbb{N}}$ in $D$ converging to a non-null vector of $\ker(\beta)$ we have that $\lim\limits_{n \to \infty} |f(x_n)| = \infty$, we can construct $w$ that satisfies $(1)$ as following.
If $f(1) = 0$ just take any $w \in f^{-1}(\mathbb{R}_{>0})$. Otherwise if $f(1) > 0$ consider $A = f^{-1}(\mathbb{R}_{>0})$ or $A = f^{-1}(\mathbb{R}_{<0})$ if $f(1) < 0$. Take any non-null element of $\ker(\beta) \cap \partial A$, say $\ell$, and a sequence $(x_n)_{n \in \mathbb{N}}$ in $A$ converging to $\ell$. Because $\lim\limits_{n \to \infty} |f(x_n)| = \infty$, there exists $N \in \mathbb{N}$ such that for every $n > N$ it follows that $f(x_n) > f(1)$. As we wanted to show.
(In fact, since without loss of generality we can substitute that sequence for a curve, there exists an uncountable number of such $w$'s.)
For example, take $\alpha(x,y) = 4x-2y$ and $\beta(x,y) = x+y$. So $f(1) = 1$. Thus
$$A = f^{-1}(\mathbb{R}_{>0}) = \{ (x+y, x-y) : x,y \in \mathbb{R}_{>0} \}$$
$$\partial A = \{ (x, x) : x \in \mathbb{R}_{\ge 0} \} \cup \{ (x, -x) : x \in \mathbb{R}_{\ge 0} \}$$
and 
$$\ker(\beta) = \{ (x, -x) : x \in \mathbb{R} \}$$
Then
$$ \ker(\beta) \cap \partial A = \{ (x, -x) : x \in \mathbb{R}_{\ge 0} \} $$
Now take $\ell = (1, -1) \in \ker(\beta) \cap \partial A$, $x_0 = (2,0)$ and $x_n = \frac{1}{n} x_0 + (1-\frac{1}{n}) \ell = (\frac{1}{n} + 1, \frac{1}{n} - 1)$. So for all $n \in \mathbb{N}$
$$f(x_n) = \frac{\frac{4}{n} + 4 - \frac{2}{n} + 2}{\frac{2}{n}} = \frac{\frac{2}{n} + 6}{\frac{2}{n}} = \frac{\frac{1}{n} + 3}{\frac{1}{n}} = 1 + 3n > 1 = f(\vec{1})$$
