A friend and I attempted to work out the proof on the board that any divergentless asymmetric tensor can be written as the sum of divergentless symmetric and antisymmetric tensors.
We wrote down the most general decomposition of the 4 by 4 tensor we could think of \begin{align} T^{mn} = \delta^{mn}h + (c^{mn}+c^{nm}-2\delta^{mn}c_{j}{}^{j}{}) + \partial^m\partial^n\phi + (\partial^mA^n+\partial^nA^m) + (\partial^mB^n-\partial^nB^m) \end{align} Computed its divergence \begin{align} \partial_mT^{mn} = \partial^nh + (\partial_mc^{mn}+\partial_mc^{nm}-2\partial^nc_{j}{}^{j}{}) + \partial^n\partial_m\partial^m\phi + (\partial_m\partial^mA^n+\partial^n\partial_mA^m) + (\partial_m\partial^mB^n-\partial^n\partial_mB^m) \end{align} and then we got stuck.
Does there exists a proof that any divergentless tensor can be decomposed into the sum of divergentless symmetric and antisymmetric tensors?
No. For a two dimensional example, take
$$ T = \begin{pmatrix} x & 0 \\ -y & 0 \end{pmatrix} $$
We have
$$ \partial_x T_{1j} + \partial_y T_{2j} = (0,0) $$
but
$$ \partial_x T_{i1} + \partial_y T_{i2} = (1,0) $$
Observe that for symmetric and antisymmetric tensors, $\partial^i T_{ij} = 0 \iff \partial^j T_{ij} = 0$. So the same is true of their sums. So the above shows that such decomposition is generally not possible.