What are all the cases in which we flip the inequality sign that don't involve multiplying by a negative? I've seen in calculus class that if we have n>N then we need to switch the inequality sign when diving 1 by both sides of the inequality, so 1/n < 1/N
It makes sense when I think about it, but I never thought I'll have to switch an inequality sign when I don't multiply by a negative, so that really surprised me.
What are other edge cases like this that they don't teach us in high school for flipping the inequality sign?
 A: Short answer: The general phenomenon of inequality switching occurs when you apply a strictly decreasing function to both sides of the equation. In your case, you are applying the function $f(x)=\frac{1}{x}$ to both sides which actually only switches the inequality some of the time: for example $2>-2$, but $\frac12>-\frac12$ (more on this later $*$).
Explanation of answer: You can think of a strictly decreasing function $f$ as a function which is always going downwards. The precise definition is that whenever $x<y$, we have $f(x)>f(y)$. Notice for example the function $g(x)=-x$ satisfies this description (and applying $g$ to both sides of an equation is equivalent to multiplying both sides by $-1$, which is the standard case of inequality switching). 
Notice that by its very definition, applying a strictly decreasing function to both sides switches the inequality.
There is perhaps an easier to see related statement: Whenever you apply a strictly $\textit{increasing}$ function to both sides, the inequality remains intact. This happens for example with the function $f(x)=x^3$.
$*$ Now let's return to the inequality switching at hand. The graph of the function $f(x)=\frac{1}{x}$ looks as follows:

You can see that this function is decreasing on the intervals $(-\infty,0)$ and $(0,\infty)$, but it is not everywhere decreasing since $f(x)<f(y)$ whenever $x<0<y$. Thus applying $\frac{1}{x}$ to both sides switches the inequality $n>N$ exactly in the cases $n,N\le 0$ or $n,N\ge0$, but maintains the inequality in the case $n>0>N$.
A: No you do not have to have a list to correctly manipulate such things! Don't even think of it this way, which will hinder your understanding and progress in math. See the reason behind it first, and then after sufficiently many times it will come naturally to you (you don't even have to memorize the "rules" at all). 
Let $n, N \geq 1$. To see why $n > N$ is equivalent to $1/n < 1/N$, you only need to apply a very basic multiplication "principle"; note that on multiplying both sides of the inequality $1/n < 1/N$ by $nN$ you get $N < n$. 
