Why do we impose de Donder gauge? In the field language, a massless particle corresponds to irreducible representations of the Lorentz group. In particular, given a spin-2 massless particle, we can embed the creation and annihilation operators into a symmetric traceless tensor $h_{\mu\nu}^T$ which satisfies
$$
\square h_{\mu\nu}^T=0\,,\qquad\partial^\mu h_{\mu\nu}^T=0\,,\qquad \eta^{\mu\nu}h_{\mu\nu}^T=0.\,\qquad \qquad (1)
$$
It is useful to consider an action for a massless spin-2 particle written in terms of a tracefull and symmetric tensor $h_{\mu\nu}$ which enjoys gauge symmetry $\delta h_{\mu\nu} = \partial_{(\mu}\xi_{\nu)}$ that is the linearised version of the Einstein-Hilbert action. In the following, I restrict to flat space-times.
In (almost) all text-books, it is used to fix the so called de Donder gauge
$$
\partial_\mu h^{\mu\nu}-\frac{1}{2}\partial^\nu h = 0.
$$
Why do we use this gauge-fixing condition? 
Why don't we impose $h=0$ and $\partial_{\mu}h^{\mu\nu}=0$ as gauge conditions separately, such that we reproduce Eq.(1) manifestly?
 A: Actually this is not possible.
To impose the gauge $h=0$, I have to solve
$$
2\partial\cdot \epsilon = -h\,,\qquad\qquad (1)
$$
which always admits solutions.
This partial gauge fixing leaves a redundant gauge freedom $\delta h_{\mu\nu}=\partial_{(\mu}\epsilon^T_{\nu)}$ where $\partial\cdot \epsilon^T=0$.
If I want to remove $\partial_\mu h^{\mu\nu}=0$, then I have to solve
$$
\square \epsilon^\mu = \partial_\nu h^{\nu\mu}\,\qquad\qquad (2)
$$
The quantity $\partial_\mu h^{\mu\nu}$ is a vector and we can decompose it into a transverse and longitudinal part
$$
\partial_\mu h^{\mu\nu} = A_\perp^\nu + \partial^\nu \pi\,\qquad\qquad (3)
$$
Then, Eq.(2) implies
$$
\square \epsilon^\mu = A^{\mu}_\perp + \partial^\nu \pi\,\qquad\qquad (4)
$$
As the parameter $\epsilon^\mu$ satisfies $\partial\cdot \epsilon=0$, the scalar field $\pi$ cannot be removed. Indeed, if I apply the divergence $\partial_\mu$ to both sides of Eq.(4) I get
$$
0 = \square \pi
$$
which is not true. Then, I am able to remove the transverse part of $\partial_\mu h^{\mu\nu}$, but not the longitudinal part.
