14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$.
14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solving, we get $b>2 \sqrt{a c}$.
Since $|O A|=\left|x_{1}\right| < 0, a+c > b$.
$$

Given that $a$, $b$, and $c$ are positive integers, we have
$$
a+c \geqslant b+1 > 2 \sqrt{a c}+1.
$$

Thus, $a+c > 2 \sqrt{a c}+1$.
Then, $(\sqrt{a}-\sqrt{c})^{2} > 1, \sqrt{a}-\sqrt{c} > 1, \sqrt{a} > \sqrt{c}+1 \geqslant 2$.
Therefore, $a > 4$, i.e., $a \geqslant 5$.
Hence, $b > 2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}=2 \sqrt{5}$, i.e., $b \geqslant 5$.
Thus, taking $a=5, b=5, c=1$, the equation $y=5 x^{2}+5 x+1$ satisfies the conditions.
Therefore, the minimum value of $a+b+c$ is 11.