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, $A+B = 2I_n$.