Let $U \subset R^n$ be a regular bounded domain having the topology of a ball. Then, the boundary value problem for $\omega\in \Omega^2(U)$,
$$
d\omega = 0
\qquad
\delta\omega = \sigma
\qquad
\mathfrak{t}(\omega) = 0
$$
is solvable provided that $\delta\sigma=0$; here $\mathfrak{t}(\omega)$ is the tangential component of $\omega$ on $\partial U$,
$$
\mathfrak{t}(\omega)(X,Y) = \omega(P_\parallel(X),P_\parallel(Y)).
$$ 
A solution exists also if the Dirichlet boundary condition is replaced by the Neumann condition $\mathfrak{n}(\omega) = 0$.

This statement remains correct if this system is for a vector-valued 2-form $\omega\in\Omega^2(U;TU)$ (solve the boundary value problem for each component $\omega^i$). Note that in this case $d$ and $\delta$ have to be replaced with the covariant exterior differential and co-differential $d^\nabla$ and $\delta^\nabla$, but the connection here is flat.

A vector-value 2-form $\omega$ can be decomposed in a collar neighborhood of $\partial U$ into a normal and a tangential component. Let $r$ be the distance from the boundary in a collar neighborhood, then we can define
$$
\omega^\perp(X,Y) = (\omega(X,Y),\partial_r)\partial_r,
$$
and $\omega^\parallel = \omega - \omega^\perp$. 

The question is the following: is the boundary-value problem with mixed boundary conditions,
$$
d^\nabla\omega = 0
\qquad
\delta^\nabla\omega = \sigma
\qquad
\mathfrak{t}(\omega^\parallel) = 0
\qquad
\mathfrak{n}(\omega^\perp) = 0
$$
solvable?

In fact, what I really need is that $\omega$ satisfies in addition $d \operatorname{tr}\omega =0$. These mixed boundary conditions guaranteed that this is indeed the case. The problem itself arises from the theory of elasticity, attempting to derive a potential theory (like Airy and Beltrami potentials) in arbitrary dimension