Answer
Yes, such a system can be consistent.
For example:
$$
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2 \\
\end{bmatrix}
$$
Work Step by Step
As long as the excess rows are linear combinations of the non-excess rows, the system is consistent. In the example above, the third equation provides no information that is not contained in the first two, but it does not contradict them either. Thus, the matrix is both overdetermined and consistent.
