Let $M$ be an internal point of the edge $AB$. Let $F$ be the midpoint of the segment $DC$.

$A_{1}$ and $B_{1}$ are the points that divide the segments $BF$ and $AF$ closer to $F$ in a 1:2 ratio, respectively.

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Since the points $A, A_{1}, M$ are in the plane $ABF$, and $MP$ is parallel to $AA_{1}$, it follows that $P$ is also in the plane $ABF$, specifically on the segment $BF$. Similarly, it can be shown that $Q$ lies on the segment $AF$.

Let $MP$ intersect the segment $BB_{1}$ at $P_{1}$, and $MQ$ intersect the segment $AA_{1}$ at $Q_{1}$.

We know that the intersection of $AA_{1}$ and $BB_{1}$ is $S$, and that $S$ is the point closer to $A_{1}$ and $B_{1}$, dividing these segments in a 1:3 ratio.

Since a reduction centered at $B$ maps $AA_{1}$ to $MP$, and a reduction centered at $A$ maps $BB_{1}$ to $MQ$, the image of $S$ under these transformations is $P_{1}$ and $Q_{1}$, respectively. Therefore, $P_{1}$ is the point closer to $P$ on the segment $MP$, and $Q_{1}$ is the point closer to $Q$ on the segment $MQ$. In other words,

$$\overrightarrow{MP} = \frac{4}{3} \overrightarrow{MP_{1}} \quad \text{and} \quad \overrightarrow{MQ} = \frac{4}{3} \overrightarrow{MQ_{1}}$$

However, $\overrightarrow{MP_{1}} + \overrightarrow{MQ_{1}} = \overrightarrow{MS}$ (since $MP_{1}SQ_{1}$ is a parallelogram), so the sum of $\overrightarrow{MP}$ and $\overrightarrow{MQ}$ is indeed $\frac{4}{3} \overrightarrow{MS}$.

If $M$ coincides with one of the endpoints of the edge $AB$, then one of the vectors $\overrightarrow{MP}$ and $\overrightarrow{MQ}$ is $\mathbf{0}$, and the other coincides with a median of the tetrahedron. Therefore, based on the properties of the centroid, the statement holds in this case as well.

Kálmán Csere (Veszprém, Lovassy L. Gymnasium IV. grade)