Suppose that a parabola has vertex $\left(\tfrac{1}{4},-\tfrac{9}{8}\right)$, and equation $y=ax^2+bx+c$, where $a>0$ and $a+b+c$ is an integer. The minimum possible value of $a$ can be written as $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
1. Given the vertex of the parabola is \(\left(\frac{1}{4}, -\frac{9}{8}\right)\), we use the vertex form of a parabola \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Here, \(h = \frac{1}{4}\) and \(k = -\frac{9}{8}\).

2. The standard form of the parabola is \(y = ax^2 + bx + c\). We need to convert the vertex form to the standard form. The vertex form is:
   \[
   y = a\left(x - \frac{1}{4}\right)^2 - \frac{9}{8}
   \]
   Expanding this, we get:
   \[
   y = a\left(x^2 - \frac{1}{2}x + \frac{1}{16}\right) - \frac{9}{8}
   \]
   \[
   y = ax^2 - \frac{a}{2}x + \frac{a}{16} - \frac{9}{8}
   \]
   Comparing this with \(y = ax^2 + bx + c\), we identify:
   \[
   b = -\frac{a}{2}, \quad c = \frac{a}{16} - \frac{9}{8}
   \]

3. We are given that \(a + b + c\) is an integer. Substituting \(b\) and \(c\) into this expression, we get:
   \[
   a + \left(-\frac{a}{2}\right) + \left(\frac{a}{16} - \frac{9}{8}\right)
   \]
   Simplifying, we have:
   \[
   a - \frac{a}{2} + \frac{a}{16} - \frac{9}{8} = \frac{16a - 8a + a}{16} - \frac{9}{8} = \frac{9a}{16} - \frac{9}{8}
   \]
   For this to be an integer, we set:
   \[
   \frac{9a}{16} - \frac{9}{8} = k \quad \text{(where \(k\) is an integer)}
   \]
   Multiplying through by 16 to clear the denominators:
   \[
   9a - 18 = 16k
   \]
   Solving for \(a\):
   \[
   9a = 16k + 18
   \]
   \[
   a = \frac{16k + 18}{9}
   \]

4. Since \(a > 0\), we need \(16k + 18 > 0\). The smallest integer \(k\) that satisfies this inequality is \(k = -1\):
   \[
   16(-1) + 18 = 2
   \]
   \[
   a = \frac{2}{9}
   \]

5. We need to express \(a\) in the form \(\frac{p}{q}\) where \(p\) and \(q\) are relatively prime. Here, \(a = \frac{2}{9}\), and \(p = 2\), \(q = 9\).

6. Therefore, \(p + q = 2 + 9 = 11\).

The final answer is \(\boxed{11}\).