2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ .
2.5.

Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$. Since $x_{1} \neq x_{2}$, we have $x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)4$.

When $a=5$, the equation $5 x^{2}-5 x+1=0$ has two distinct positive roots in the interval $(0,1)$. Therefore, the minimum value of $a$ is 5.