Why does the first lithium stationary state have n = 9/4?
I think your calculation is correct. There is an error in this textbook.  All you have to do is to switch  between $\ce{Li^{2+}}$ and  $\ce{He^{+}}$. And 'll prove my point:  

$\ce{Li^{2+}}$ and  $\ce{He^{+}}$ are hydrogen-like atoms.

For these species, the ionization energy is given by the equation:
$$E_\mathrm{i}=E_\mathrm{i}(\ce{H})\times Z^2$$

For $\ce{Li^{2+}}$: 
\begin{align}
E_\mathrm{i}(\ce{Li^{2+}})&amp;=E_\mathrm{i}(\ce{H})\times Z^2\\
19.6 \times 10^{-18}&amp;=E_\mathrm{i}(\ce{H})\times 3^2\\
E_\mathrm{i}(\ce{H})&amp;=2.18 \times 10^{-18}  \,\mathrm {J/atom}\\
E_\mathrm{i}(\ce{H})&amp;=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\
  &amp;= 13.6 \,\mathrm{eV}
\end{align}

$13.6 \,\mathrm{eV}$ is well-known to be the first ionization energy of hydrogen. So, my assumption is true.
To solve the problem after correction:
 The ionization energy of $\ce{He^{+}}$ is given by the equation:
\begin{align}
E_\mathrm{i}(\ce{He^{+}})&amp;=\frac{E_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\
E_\mathrm{i}(\ce{He^{+}})&amp;=8.72\times 10^{-18}\,\mathrm {J/atom}
\end{align}
The energy of the first stationary state:
$$E_\mathrm{i,1}=E_\infty-E_1$$
But, $E_\infty=0$, we find:
$$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$