jmc

Since the vertex is at $\left(\frac{1}{4},-\frac{9}{8}\right)$, the equation of the parabola can be expressed in the form $$y=a{\left(x-\frac{1}{4}\right)}^2-\frac{9}{8}.$$Expanding, we find that $$y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8}=a{x}^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.$$From the problem, we know that the parabola can be expressed in the form $y=a{x}^2+bx+c$, where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$, $b=-\frac{a}{2}$, and $c=\frac{a}{16}-\frac{9}{8}$. Adding up all of these gives us $$a+b+c=\frac{9a-18}{16}=\frac{9(a-2)}{16}.$$Let $n=a+b+c.$ Then $\frac{9(a-2)}{16}=n,$ so $$a=\frac{16n+18}{9}.$$For $a$ to be positive, we must have $16n+18>0,$ or $n>-\frac{9}{8}.$ Setting $n=-1,$ we get $a=\frac{2}{9}.$

Thus, the smallest possible value of $a$ is $\boxed{\frac{2}{9}}.$