Einstein notation: can a free index be upper in one term and lower in another term? Consider a linear combination of terms written using Einstein notation. 
Consider one free index in the linear combination: is it necessary that the index is upper in all terms or lower in all terms, or not?
Equivalently is it possible to have the same free index, but in one term it is a upper index and in the other it is a lower index?
For example, is it possible to have the following expression?
$$A_\mu+B^\mu$$
Or this is forbidden in einstein notation, and one free index can only appear upper in all terms or lower in all terms?
 A: Short answer: No.
When dealing with contravariant vectors (also just vectors $\rightarrow$ upper indices) and covariant vectors (also covectors $\rightarrow$ lower indices), you can only sum vectors that transform the same way with respect to change of basis.
Think of a vector as a column vector and a covector as a row vector, to make this simple to assimilate.
Also, a contravariant vector can only be contracted with a contravariant vector, corresponding to the summation of the products of their coefficients. On the other hand, when there is a fixed coordinate basis (or when not considering coordinate vectors), one may choose to use only subscripts.
A: Well, you can write down the expression if you want to, and calculate its value, but there is a problem: it doesn't have any well defined transformation law when you change coordinates. Say you switch to a different frame through a change-of-basis matrix $C^\mu{}_\nu$. The components of a contravariant vector $A^\mu$ change to $C^\mu{}_\nu A^\nu$ and those of a covector $B_\mu$ change to $B_\nu (C^{-1})^\nu{}_\mu$, so if you form a combination like
$$D^\mu = A^\mu + B^\mu,$$
in a new basis we have
$$D'^\mu = C^\mu{}_\nu A^\mu + C^\mu{}_\nu B^\mu = C^\mu{}_\nu (A^\nu + B^\nu)= C^\mu{}_\nu D^\nu,\tag{1}$$
so its components will also change like a vector. But under a change of basis, the expression $A^\mu + B_\mu$ changes to
$$C^\mu{}_\nu A^\mu + (C^{-1})^\nu{}_\mu B_\nu$$
which does not obey any simple transformation law. In particular, and this is the important part, the way it transforms depends on the values of $A^\mu$ and $B_\mu$ individually instead of simply depending on $A^\mu + B_\mu$; compare with $(1)$ above, where the transformation just depends on $D^\mu$.
It's for this reason that the notation doesn't let you give it a name: both $D^\mu = A^\mu + B_\mu$ and $D_\mu = A^\mu + B_\mu$ would be wrong, because the position of the index on $D$ is in both cases implying a transformation law that is not the one $D$ actually obeys.
