14. Let $a$, $b$, $c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
14. Let the two real roots of the equation be $x_{1}, x_{2}$. By Vieta's formulas, we know that $x_{1}, x_{2}$ are both negative.
From $\frac{c}{a}=x_{1} x_{2}9$. Therefore, $b^{2} \geqslant 4 a c=4 \times \frac{a}{c} \times c^{2}>4 \times 9 \times 1^{2}=36$.
Solving this, we get $b>6$. Hence, $b \geqslant 7$.
Also, $\frac{b}{a}=\left(-x_{1}\right)+\left(-x_{2}\right)\frac{3}{2}, a>\frac{3}{2} b \geqslant \frac{21}{2}$. Therefore, $a \geqslant 11$.
(1) When $b=7$, from $4 a \leqslant 4 a c \leqslant b^{2}$ and $a \geqslant 11$, we know $a=11$ or $12, c=1$. However, the equation $11 x^{2}+7 x+1=0$ has roots $-\frac{7+\sqrt{5}}{22}-\frac{1}{3}$, solving this gives $4+\sqrt{16-a}0$.

And $f(15)=0$, so $a$ can only be 16.
At this point, $a+b+c=16+8+1=25$, and the roots of $16 x^{2}+8 x+1=0$ are $x_{1}=x_{2}=-\frac{1}{4}$, which satisfies the conditions.
(3) When $b \geqslant 9$, $a>\frac{3}{2} b \geqslant \frac{27}{2}$, so, $a \geqslant 14$. Thus, $a+b+c=14+9+1 \geqslant 24$.

If $a+b+c<25$, then only $a=14, b=9, c=1$ is possible. In this case, the roots of the equation $14 x^{2}+9 x+1=0$ are $x_{1}=-\frac{1}{2}, x_{2}=-\frac{1}{7}$, which does not meet the conditions. Therefore, in this case, $a+b+c \geqslant 25$.
In summary, the minimum value of $a+b+c$ is 25.