Yes, the mean squared error (MSE) and mean absolute error (MAE) can be equal.

Let $r_i = |y_i - \hat{y}_i|$ denote the absolute value of the residual for the $i$th data point, and let $r = [r_i, \dots, r_n]^T$ be a vector containing absolute residuals for all $n$ points in the dataset. Letting $\vec{1}$ denote a $n \times 1$ vector of ones, the MSE and MAE can be written as:

$$MSE = \frac{1}{n} r^T r \quad
MAE = \frac{1}{n} \vec{1}^T r
\tag{1}$$

Setting the MSE equal to the MAE and rearranging gives:

$$(r - \vec{1})^T r = 0 \tag{2}$$

The MSE and MAE are equal for all datasets where the absolute residuals solve the above equation. Two obvious solutions are: $r = \vec{0}$ (there's zero error) and $r = \vec{1}$ (the residuals are all $\pm 1$, as mkt mentioned). But, there are infinitely many solutions.

We can interpret equation $(2)$ geometrically as follows: The LHS is the dot product of $r-\vec{1}$ and $r$. Zero dot product implies orthogonality. So, the MSE and MAE are equal if subtracting 1 from each absolute residual gives a vector that's orthogonal to the original absolute residuals.

Furthermore, by completing the square, equation $(2)$ can be rewritten as:

$$\Big( r-\frac{1}{2} \vec{1} \Big)^T \Big( r-\frac{1}{2} \vec{1} \Big)
= \frac{n}{4} \tag{3}$$

This equation describes an $n$-dimensional sphere centered at $[\frac{1}{2}, \dots, \frac{1}{2}]^T$ with radius $\frac{1}{2} \sqrt{n}$. The MSE and MAE are equal if and only if the absolute residuals lie on the surface of this hypersphere.