Entanglement of formation of the mixture of maximally entangled states Suppose we have two spin-$S$ systems. Let $\left| \psi_{a,b} \right\rangle = \frac{1}{\sqrt{2}}(\left| a,b \right\rangle+\left| b,a \right\rangle)$ be the maximally entangled state. ($a\neq b$ and $-S\leq a,b \leq S$.) 
What is the Entanglement of formation for $\rho=\frac{1}{C} \sum_{a\neq b} \left| \psi_{a,b} \right\rangle \left\langle \psi_{a,b} \right|$? $C$ is the normalizing constant.
A trivial upper bound is 1. But can we give a nontrivial upper bound or even calculate it explicitly?
 A: Since you say that any other entanglement measure is fine, let's compute the negativity.  Let me denote by $\rho_{ab}:=|\psi_{a,b}\rangle\langle\psi_{a,b}|$.
With $^{T_A}$ the partial transpose, we have
$$
\rho_{ab}^{T_A} = \frac12 \Big[
|a,a\rangle\langle b,b|+|b,b\rangle\langle a,a|+
|a,b\rangle\langle a,b|+|b,a\rangle\langle b,a|\Big]\ .
$$
Thus (denoting by $D:=2S+1$ the number of basis states),
$$
\rho^{T_A} = \frac{2}{D(D-1)}\sum_{a>b} \rho_{ab}^{T_A}
$$
is block-diagonal with two blocks: $\rho^{T_A}_{ab,a'b'}$ for $a\ne b$, $a'\ne b'$ is diagonal with entries $\tfrac{1}{D(D-1)}$ (i.e., $D(D-1)$ entries), and  $\rho^{T_A}_{aa,a'a'}$ (a $D\times D$ matrix) equals $\tfrac{1}{D(D-1)}$ everywhere except on the diagonal (which is zero). Since the latter equals 
$$
\tfrac{D}{D(D-1)}|+\rangle\langle +|-\tfrac{1}{D(D-1)}1\!\!1\ ,
$$
with $|+\rangle = (\sum |a\rangle)/\sqrt{D}$, it has eigenvalues $-\tfrac{1}{D(D-1)}$ with multiplicity $D-1$ and $\tfrac{1}{D}$ with multiplicity $1$, respectively.
The sum of the absolute value of the eigenvalues of $\rho^{T_A}$ is thus
$$
\|\rho^{T_A}\|_1={D(D-1)}\frac{1}{D(D-1)}+(D-1)\frac{1}{D(D-1)}+\frac{1}{D} = 
1+\frac{2}{D}\ .
$$
The negativity is thus
$$
\mathcal N(\rho) = \frac{\|\rho^{T_A}\|_1-1}{2} = \frac{1}{D}
$$
and the log-negativity
$$
E_N(\rho) = \log(\|\rho^{T_A}\|_1) = \log(1+2/D)\ .
$$
