All possible values of $n$ and $m$ are $(n, m)=(2k, 2l)$ for some $k, l\in\mathbb N$. Construction: color the rectangle floor mats like this: $$\begin{array}{|c|c|c|c|c|c|c|}\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline \end{array}$$Put babies facing right on $R$, facing down on $D$, facing left on $L$, facing up on $U$. One can see that after one clap, no babies will bump into each other or crawl outside of the rectangle, and the facing of the babies on $R$ are still right, $D$ are still down, $L$ are still left, $U$ are still up. $\therefore$ no matter how many times the nanny claps, for any mat, the facing of the baby on it is always the same, and therefore no baby would cry. For $(n, m)\neq(2k, 2l)$, color the rectangle floor mats similarly to the above. $$\begin{array}{|c|c|c|c|c|c|c|}\hline R&D&R&D&\cdots\\\hline U&L&U&L&\cdots\\\hline R&D&R&D&\cdots\\\hline U&L&U&L&\cdots\\\hline \vdots&\vdots&\vdots&\vdots&\ddots\\\hline \end{array}$$Suppose the opposition that no baby would cry no matter how many times the nanny claps. One can see that if $2\nmid n$ or $2\nmid m$, the number of $R>$ the number of $L$, and after two claps, all babies on $R$ will move to $L$. However, since the number of $R>$ the number of $L$, there exists two babies that will bump into each other, contradiction. This finishes the proof.

All possible values of $n$ and $m$ are $(n, m)=(2k, 2l)$ for some $k, l\in\mathbb N$. Construction: color the rectangle floor mats like this: $$\begin{array}{|c|c|c|c|c|c|c|}\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\hline R&D&R&D&\cdots&R&D\\\hline U&L&U&L&\cdots&U&L\\\hline \end{array}$$Put babies facing right on $R$, facing down on $D$, facing left on $L$, facing up on $U$. One can see that after one clap, no babies will bump into each other or crawl outside of the rectangle, and the facing of the babies on $R$ are still right, $D$ are still down, $L$ are still left, $U$ are still up. $\therefore$ no matter how many times the nanny claps, for any mat, the facing of the baby on it is always the same, and therefore no baby would cry. For $(n, m)\neq(2k, 2l)$, color the rectangle floor mats similarly to the above. $$\begin{array}{|c|c|c|c|c|c|c|}\hline R&D&R&D&\cdots\\\hline U&L&U&L&\cdots\\\hline R&D&R&D&\cdots\\\hline U&L&U&L&\cdots\\\hline \vdots&\vdots&\vdots&\vdots&\ddots\\\hline \end{array}$$Suppose the opposition that no baby would cry no matter how many times the nanny claps. One can see that if $2\nmid n$ or $2\nmid m$, the number of $R>$ the number of $L$, and after two claps, all babies on $R$ will move to $L$. However, since the number of $R>$ the number of $L$, there exists two babies that will bump into each other, contradiction. This finishes the proof.