Write $x=\overline{ab}$ and $y=\overline{cd}$, then the condition is $xy\mid 100x+y$, so $x\mid y$. Let $y=kx$, with $k\in \mathbb{N}$, then the condition is $kx^2\mid 100x+kx$, so $k\mid 100$. Size reasons give $k=2$, $k=4$ or $k=5$ as the only possibilities. If $k=2$ we get $2x^2\mid 100x+2x$, so $x\mid 51$, which gives $x=17$ (because $x\geq 10$ and $y<100$). If $k=4$ we get $4x^2\mid 100x+4x$, so $x\mid 26$, which gives $x=13$ (because $x\geq 10$ and $y<100$). If $k=5$ we get $5x^2\mid 100x+5x$, so $x\mid 21$, which gives no solutions (because $x\geq 10$ and $y<100$). Hence the guanaco numbers are $1352$ and $1734$ (it's clear that both work).

Nice problem too! My solution is similar with #2 (is it possible to imagine something too another?)

This problem appeared on the Crux Matemathicum Journal in 2011 link: https://cms.math.ca/wp-content/uploads/crux-pdfs/CRUXv37n7.pdf it's on page 437.