I assume that what the OP wanted to say was, given a correlation matrix $B$, find a correlation matrix $A$ that maximizes $\det(A+B)$. Let me cite here a more general theorem that paves the way to a solution.

> **Theorem** Let $A$ and $B$ be Hermitian matrices with eigenvalues $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$, respectively. Then, 

\begin{equation*}
 \det(A+B) \le \max_{\sigma \in \mathfrak{S}_n}\prod_{i=1}^n (a_i + b_{\sigma(i)}).
\end{equation*}

Since correlation matrices are Hermitian positive semidefinite, a specialization of this theorem shows us that
\begin{equation*}
\det(A+B) \le \prod_{i=1}^n (\lambda_i(A) + \lambda_{n-i+1}(B)),
\end{equation*}
where $\lambda_i(\cdot)$ is the $i$-largest eigenvalue. Thus, in particular, for a fixed $B$, we just need to pick a suitable $A$ that has the same eigenvectors as $B$ (in permuted order, however), but has the largest possible allowable eigenvalues. 

*EDIT* It is not immediate, if these eigenvalues can be easily obtained in closed form. It seems that the optimum solution is obtained by setting
\begin{equation*}
a_{ij} = \begin{cases}
 -b_{ij} & i \neq j,\\
 1       & i = j
 \end{cases}.
\end{equation*}
In this case, $A+B = 2I_n$.