Given a triangle $ABC$, and a point $D$ on the edge $BC$, is it possible to express vector $\overrightarrow{AD}$ with positive integers $x,y$ and vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ such that: $$\overrightarrow{AD} = \frac{x\overrightarrow{AB}+y\overrightarrow{AC}}{x+y}$$
This was an explanation used in Solution $3$ in the 2019 AIME Problem $4$, but I can't seem to find any justification or proof for the claim.
Any help would be appreciated.
The solution doesn't claim that $ x $ and $ y $ are positive integers in the beginning, but only positive (real numbers). The fact that we can find such integers is only shown later, and is because the other ratios given in the question are rational numbers, and rational numbers are closed under addition, subtraction, multiplication and division (by nonzero rationals). But if your question is also about why there are such positive reals $ x $ and $ y $, the following may help.
Consider any point $ D $ on the segment $ B C $. Let $ x $ be the length of the segment $ C D $ and $ y $ be length of the segment $ B D $, so that the length of $ B C $ is $ x + y $ (and thus $ x + y > 0 $). We have $ 0 \le x , y \le x + y $ (with either $ x > 0 $ or $ y > 0 $), and in case $ D $ is strictly between $ B $ and $ C $, we have $ 0 < x , y < x + y $. It's then fairly easy to see that $$ \overrightarrow { B D } = \frac y { x + y } \overrightarrow { B C } \quad \text {and} \quad \overrightarrow { C D } = \frac x { x + y } \overrightarrow { C B } \text . $$ The rest is simple algebraic manipulation of vectors: $$ x \overrightarrow { B D } + y \overrightarrow { C D } = \frac { x y } { x + y } \left( \overrightarrow { B C } + \overrightarrow { C B } \right) = \vec 0 \text ; $$ $$ \therefore \ x \left( \overrightarrow { A D } - \overrightarrow { A B } \right) + y \left( \overrightarrow { A D } - \overrightarrow { A C } \right) = \vec 0 \text ; $$ $$ \therefore \ ( x + y ) \overrightarrow { A D } = x \overrightarrow { A B } + y \overrightarrow { A C } \text ; $$ $$ \therefore \ \overrightarrow { A D } = \frac { x \overrightarrow { A B } + y \overrightarrow { A C } } { x + y } \text . $$ Note that we could get the same results if we took $ x $ and $ y $ as not necessarily the length of $ C D $ and $ B D $, but $ \alpha $ times each of the lengths, for some positive real $ \alpha $. This shows that if the ratio between the lengths of the segments is a rational number, we can take $ x $ and $ y $ to be integers (since either $ x > 0 $ or $ y > 0 $, at least one of the ratios $ \frac y x $ and $ \frac x y $ can be defined).