6. Given the quadratic function $f(x)=a x^{2}+b x+c, a$ $\in \mathbf{N}_{+}, c \geqslant 1, a+b+c \geqslant 1$, the equation $a x^{2}+b x+c$ $=0$ has two distinct positive roots less than 1. Then the minimum value of $a$ is
6.5.

Let the two roots of the equation $a x^{2}+b x+c=0$ be $x_{1}$ and $x_{2}$, and $0<x_{1}<x_{2}<1, a>4$.
Therefore, $a \geqslant 5$.
Take $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)$, which meets the conditions. Hence, the minimum value of $a$ is 5.