For the LRC loop rule, why is the $-L \frac{di}{dt}$ term negative?

The signs can be a bit confusing in these loops, as currents can flow in either direction and the quantities can be increasing or decreasing, so the derivatives of charges and currents can be positive or negative. The important thing is that the relation between the signs in the resulting differential equation is right -- the solution should then give you the right actual signs of currents and voltages. I wrote down $-L \frac{di}{dt}-iR + q/C = 0$, and then $L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0$, as follows:

Imagine the capacitor is has just been charged up (say, $+q$ on the top plate, $-q$ on the bottom) and take the moment at which the current has just started flowing counter clockwise (the loop rule will still be true at any moment). We'll take $q$ to be the charge on the top plate. Step around the loop in the direction of the current. The voltage change across the capacitor is $+q/C$, and the voltage change across the resistor in the direction of the current is $-iR$. The inductor will be developing an EMF proportional to $L\frac{di}{dt}$. If $i$ is CCW as the discharge starts, the inductor (by Lenz's Law) is creating a back-emf to counteract the increase in current, so an emf that is negative, $-L \frac{di}{dt}$. So we have $-L \frac{di}{dt}-iR + q/C = 0$. But then, since $q$ is the charge on the top plate, and it's decreasing, we should write $i= -\frac{dq}{dt}$ and so $L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0$.

Similarly, for the LC circuit we write $- L \frac{di}{dt} + \frac{q}{C} = 0$, then take $i= -\frac{dq}{dt}$ to get $L \frac{d^2 q}{dt^2} +\frac{q}{C} = 0$.