The correct answer is $\boxed{\text{C. n distinct pair of complex conjugate numbers}}$.
A square matrix $A$ is said to be symmetric if $A^T = A$, where $A^T$ is the transpose of $A$. If $A$ is a real symmetric matrix, then its eigenvalues are always real. Moreover, if $A$ is a square matrix of dimensions $2n$, then its eigenvalues come in complex conjugate pairs. This is because if $\lambda$ is an eigenvalue of $A$, then so is $\overline{\lambda}$, where $\overline{\lambda}$ is the complex conjugate of $\lambda$.
To see why this is the case, let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Then, $Av = \lambda v$. Taking the transpose of both sides, we get $v^T A^T = \lambda v^T$. Since $A^T = A$, this gives $v^T A = \lambda v^T$. Multiplying both sides by $\overline{\lambda}$, we get $\overline{\lambda} v^T A = \overline{\lambda} v^T$. Taking the transpose of both sides again, we get $A \overline{v^T} = \overline{\lambda} \overline{v^T}$. Since $A$ is real, this gives $A \overline{v^T} = \overline{v^T} \lambda$. Hence, $\overline{v^T}$ is also an eigenvector of $A$ corresponding to the eigenvalue $\overline{\lambda}$.
Therefore, if $A$ is a real symmetric matrix of dimensions $2n$, then its eigenvalues come in complex conjugate pairs. In other words, if $\lambda$ is an eigenvalue of $A$, then so is $\overline{\lambda}$.