One, (20 points) Find all positive real numbers $a$ such that the equation $x^{2}-a x+4 a=0$ has only integer roots.

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Let $\Delta=(-a)^{2}-4 \times 4 a \geqslant 0$, and $a>0$, then $a \geqslant 16$.
Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-a x+4 a=0$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=a, x_{1} x_{2}=4 a \text{. }
$$

Thus, $x_{1} x_{2}-4 x_{1}-4 x_{2}=0$,
or $\left(x_{1}-4\right)\left(x_{2}-4\right)=16$.
Considering the symmetry of $x_{1}, x_{2}$ and that they are integers, we have
$$
\begin{array}{c}
\left\{\begin{array}{l}
x_{1}-4= \pm 16, \pm 8, \pm 4, \\
x_{2}-4= \pm 1, \pm 2, \pm 4 .
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
x_{1}=20,12,8,-12,-4,0, \\
x_{2}=5,6,8,3,2,0 .
\end{array}\right.
\end{array}
$$

But $a \geqslant 16$,
thus, $a=25,18,16$.