First, we may pull out powers of $X$ in both power series to get series starting at $0$:
$P(X) = \sum_{n\geq r} p_n X^n = X^r \cdot \sum_{n \geq 0} p_{r+n} X^n \\ Q(X) = \sum_{n\geq s} q_n X^n = X^s \cdot \sum_{n \geq 0} q_{s+n} X^n.$
Now, to get a solution $A(X)$ to $Q(X) A(X) = P(X)$ we need to be able to find a solution $A(X) = P(X) / Q(X)$. In terms of the above series, this means
$A(X) = \frac{P(X)}{Q(X)} = \frac{X^r \cdot \sum_{n \geq 0} p_{r+n} X^n}{X^s \cdot \sum_{n \geq 0} q_{s+n} X^n} = X^{r-s} \cdot \frac{\sum_{n \geq 0} p_{r+n} X^n}{\sum_{n \geq 0} q_{s+n} X^n}.$
Since $q_s \neq 0$ the series $\sum_{n \geq 0} q_{s+n} X^n$ is invertible. So all we need is that the powers of $X$ are all non-negative. From $p_r \neq 0$ it follows that we need $r - s \geq 0$ or $\boxed{r \geq s}$. This condition is also sufficient: if $r \geq s$, then the unique solution $A(X)$ to the equation is given by
$A(X) = X^{r-s} \cdot \left(\sum_{n \geq 0} p_{r+n} X^n\right) \cdot \left(\sum_{n \geq 0} q_{s+n} X^n\right)^{-1}.$
If you want, you can even find somewhat explicit formulas for the coefficients $a_i$ of $A(X)$, i.e., $\begin{align} a_{n} &= 0 \qquad \qquad (n = 0, \ldots, r-s-1) \\ a_{r-s} &= \frac{p_r}{q_s} \\ a_{r-s+1} &= \frac{p_r}{q_s}\left(1 - \frac{q_{s+1}}{q_s}\right) \\ a_{r-s+2} &= \frac{p_r}{q_s}\left(1 - \frac{q_{s+1}}{q_s} - \frac{q_{s+2}}{q_s} + \frac{q_{s+1}^2}{q_s^2}\right) \\ \vdots \quad &= \quad \vdots \end{align}$
Note that these terms only contain divisions by $q_s$, which is ok, since we know $q_s \neq 0$.