How many moles NH4Cl must be added to NH3 to create buffer with pH=9?
$\ce{NH3}$ is such a weak base ($K_{\mathrm{b}}=1.8\times 10^{-5}$) that the initial concentration of $\ce{NH4+}$ can be considered negligible. You can consider the initial concentration of $\ce{NH4+}$, but the answer you will get is the same to a considerable number of significant figures.
By considering the equilibrium of $\ce{NH3}$ dissociating to form $\ce{NH4+}$ and $\ce{OH-}$ it is easy to find the initial concentration of $\ce{NH4+}$.
$$K_{\mathrm{b}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$
Since \begin{align}[\ce{NH4+}] &amp;= [\ce{OH-}] \tag{1}\\
K_{\mathrm{b}}\cdot [\ce{NH3}] &amp;= [\ce{NH4+}]^{2} \tag{2}
\end{align}
Plugging in values gives:
$$[\ce{NH4+}] = 1.34\times 10^{-3} (3\text{s.f.})$$
Which equates to $\pu{2.68\times 10^{-3} mol}$ in the $\pu{2 L}$ solution.
If I'm right this should be orders of magnitude smaller than the textbook answer, and therefore negligible.