Large resistance of diodes measured by ohmmeters
The open-circuit voltage from a modern digital multimeter in ohms mode is not very well defined - I measured a few and they varied from less than 0.6V to more than 2V. 

Most of them will allow a silicon diode or schottky diode or germanium diode to conduct somewhat so on some range you could get a polarity indication from a common type of diode (though probably not an LED). 

The diode function typically puts around 1-2mA through the probes with a compliance of more than 2.5V so it gives you some indication of the forward voltage of diode (and maybe a red or IR LED) at a sensible current. 

Your 1N34A is, of course, an ancient germanium diode and has a relatively low forward voltage. Maybe 300mV at 1mA, not 600mV. That drop increases rapidly at higher currents. If you use the diode mode you should be able to see the difference between a 1N34A and, say, a 1N914/1N4148 readily, and a Schottky such as BAT54 or 1N5817/19 will be easily distinguished from, say, a UF4007. 



You are correct that the incremental forward resistance of a puny germanium diode is of the order of a hundred ohms at mA-ish currents (order of magnitude) but remember that the ohmmeter is not measuring incremental resistance it is measuring the total voltage drop across the nonlinear device and dividing that by that current I mentioned (measured typically by deriving the ADC reference from a dropping resistor) to give you the equivalent total resistance.

For example, a 'stiff' voltage source of 0.3V (0\$\Omega\$ incremental resistance) would give you a large resistance reading on some range (with the right polarity). (obviously you shouldn't be putting volts into a meter on ohms range generally, but that is what would happen). 

As a reminder, the dynamic resistance of an ideal (Shockley model) diode at current I is: 

r = \$\frac{\eta V_T}{I}\$ 

where \$\eta\$ is the ideality factor (between 1 and 2 - ~=1.3 for a 1N34A diode) and I is the diode current, and \$V_T\$ is the thermal voltage - 26mV at room temperature. 

So at 1mA r ~= 34\$\Omega\$

To find the 'resistance' a meter would measure- let's assume it's on a 20K range and the meter measuring current is about 50uA (what one of mine uses), and \$\eta\$ = 1.3, Is = 200nA (from a 1N34A SPICE model).  

\$R_{EQ} = \frac{\eta V_T \ln(I/I_S)}{I}\$= 3.7k\$\Omega\$

Since your meter is reading a little higher, probably the current is lower than 50uA- maybe 30uA. You can measure  approximately by using another meter in current mode. 

By comparison, the dynamic resistance at 50uA is less than 700\$\Omega\$, so a huge difference.



The magnitude of the 'resistance' readings you get from a diode in resistance mode are basically meaningless unless  you know the internal workings of the multimeter, so all you can divine is that the diode conducts to some reasonable degree in that particular direction. As you have observed they will change drastically from on resistance range to another. In diode mode you usually get a reading that's indicative of a forward voltage drop at a mA or two roughly, which is infinitely more useful. They don't control the current well, they often don't document it well, it's just a free add-on function, but handy none the less. 



By the way, almost all digital multimeters (on ohms mode) have the red lead as the positive output in diode/resistance mode but older analog meters often reversed that.