an/prm-10 grid dip question - how the voltage can be negative at the grid?
I was studying tube designs as a child, long before I had any clue how a grounded cathode amplifier might be designed. Which means I don't remember that much. But one thing I do recall, which had bothered me a lot back then, is the ability of a grid to self-bias. You might look up "grid leak resistor," or something similar.

A vacuum tube's cathode may either be the heated filament itself, or else the heated filament can be thermally connected to an electrically isolated plate. Either way, heat causes electrons to sputter off the surface of the cathode and form a nearby cloud of sorts.

(Electrons can also be ripped off using a high voltage, but that's a different process. If interested, look up the term "Wehnelt" where a combination of thermal and field effects are used. See how it applies to electron beams used in electron microscopes or electron beam lithography.)

With the plate present and positively charged relative to the cathode, this cloud of electrons is accelerated towards the plate. (The electrons will shoot across the interstitial vacuum towards the plate.) Or would be, except for the grid. A grid (or screen) is placed in between the two and used to control the acceleration, along the way.

If the grid is biased externally, it can be either positive or negative. If negative, it hinders the ability of the electrons to "see" the plate's positive charge a bit and retards their acceleration at first. For those electrons that cross through the grid, the negative grid then acts to add acceleration to them and propels them with even more acceleration. If positive, it adds some acceleration to the electrons prior to their crossing through the grid area. And whether positive or negative, a varying grid voltage has a varying impact on the plate current.

If the grid were nothing more than affecting the accelerating potential that the electrons experience in a vacuum, then all of the cathode current would equal all of the plate current and that would be that. (Like an ideal MOSFET.) However, it's in the middle of things and a few of the electrons smack into it, as well. So if the grid is left floating then electrons accumulating onto the grid will either charge up the grid voltage, adding to its negative field (until the electron cloud either no longer moves or else in steady state simply moves entirely around the grid, in effect), or else there must be some mechanism provided to remove them.

One way of doing this is by adding what is called a "grid leak resistor" from the grid. If all you do is hook up a resistor from the grid to ground (or somewhere that can accept them), then the grid will rapidly self-bias negatively; until the leakage through the resistor due to the voltage on the grid is sufficient to match the addition of impacting electrons on the grid due to the cathode to plate current and it reaches a steady state voltage. Using a grid leak resistor in this way is called "self-biasing" and it avoids the need for a separate power supply.

In these one-tube triode wonders, the whole thing was just called a "grid leak detector." You can look that up, as well. They'd also often use the grid leak detector to directly drive headphones, too. If you look at the headphone jack, you can see that it grounds out a node at the other end of \$R_{204}\$ when not in use but that a headphone would be inserted in between, otherwise.

In any case, I'm not surprised that you see a slightly negative voltage there, without seeing any negative voltage developed as the result of a separate power supply rail. I can't necessarily explain why you see a positive voltage when reversing the leads of your meter, though. (That's a different issue.) Also, the grid does include a weak AC signal superimposed upon this self-biasing DC grid voltage. So your comment saying that "the voltage at pin 3 should be -10V AC" seems fine to me, if interpreted to mean "an AC signal added to an expected DC bias of \$-10\:\textrm{V}\$."