There are not good answers to these questions because LEDs are intended for emitting light, and as such the parameters you need to answer your questions are not specified.
A LED reverse biased as a light sensor is a current source proportional to the light level. Being a current source, it have very high impedance (a perfect current source has infinite impedance). The response time is proportional to the resistance of the node times the capacitance. Since the capacitance is parasitic, it is hard to guess and will depend a lot on the particular LED and on layout. The resistance is the deliberate resistance R1 in parallel with any leakage resistance and the resistance of the LED being a imperfect current source. Other then R1, these are again hard to guess. 20 MΩ is so high that leakage can be a important factor. Even dirt on the board and ambient humidity matters at that impedance.
As for how to determine the voltage, that again must be done experimentally. Unless you have a unusual LED that is intended also for reverse operation, you're not going to get a spec. Test a few and leave lots of room for device variation.
I would use a considerably lower resistance with some amplification. The lower resistance will decrease the response time and make things more predictable by making the leakage resistance small enough in comparison to not matter. You are currently getting ouputs from 150 mV to 5 V with 20 MΩ. With 2 MΩ instead, those voltages will be 15 mV to 500 mV, which is still big enough for plenty of opamps to amplify reliably and should be low enough to make leakage ignorable. It may still be too slow, in which case you can use lower resistance still with better amplification.
Another point is that if your supply is large enough to get 5V on R1, then you may be applying too much reverse voltage to the LED in low light conditions. Check the LED datasheet (this usually is specified) and make sure you're not exceeding the reverse voltage limit. A lower resistance will let you use lower reverse bias voltage.
Some answers have explained how to 'multiplex' the emitters and sensors. The scheme suggested here can use that, or not. For emitters there is no need if the sensors are well isolated from the emitters.
You seem to be very cost conscious, as am I. The cheapest way to make a simple reflective sensor is with an InfraRed (IR) emitter and IR phototransistor. That pair will be well under 0.20GBP, which is much lower cost than the TCRT5000.
For example 940nm 5mm IR emitter 940nm 5mm IR phototransistor. Most distributers should have parts in this price range. To give some context, these would cost about £4.60 for 100 IR emitters, and £7.50 for 100 phototransistors. These are Farnell prices, which is one of the most expensive distributers. So you should be able to do much better.
You could drive the IR emitters with a chain of shift registers. Use 'Serial In Parallel Out' (SIPO) shift registers. They will take a serial data pattern to choose which emitters to drive. That would consume three or four pins (clock, data, enable, latch) for all 100 emitters. For example TLC5916
If the internal structure of the board shields IR phototransistors from IR emitters you could even drive many emitters simultaneously using each pin of a chip like the TLC5916 which would drive much more current than an Arduino pin. So you might be able to drive all of the emitters with a couple of TLC5916.
With a bit of experiment, you should be able to read all of the sensors using simple logic-level shift-registers too. Parallel In, Serial Out (PISO) shift-registers. The issue is not power, so cheaper parts are sufficient, for example a 74HC165 (which should be available for under $1). These can be 'chained' end to end, and even used with a multiplexing scheme.
Daylight, or electric lighting may interfere, and 'confuse' the IR sensors. One way to cope with that is to 'modulate' the IR LED/emitter. Using one LED+sensor as an example: switch the IR LED on, take a measurement from the sensor, switch the LED off, and take another. When the difference is very small, there is a lot of light interference, and very little reflected light, so the piece is unlikely on the square. When the difference is large, the piece is likely on the square because there is very little light interference, and the reflected light is strong.
Experiments might demonstrate that this is not an issue. However, having the flexibility to deal with this type of interference might become very important. So try to ensure it is practical to with IR LED/emitter on and off, until you have some real evidence to show it is not an issue.
Best Answer
I repair guns for a system called "Battle Sports". That employs an infra-red emitter combined with a couple of white LEDs in the gun:
That module is placed at one end of a plastic tube, about 6 inches long. At the other end of the tube is a lens. The lens is specifically shaped, and placed, so that the light from the module is focused into a roughly parallel beam. That translates to the white light forming a spot a couple of inches across (used to aim with), and the IR light covering roughly the same area.
The IR is modulated at 56KHz, with a pulsed on/off signal to identify that it's a proper gun that's in use.
The gun also has an IR receiver module on it, as well as a pair of receivers that the user wears on a headband. These are a simple amplifier / filter arrangement that just returns the pulses without the 56KHz modulation. They are domed, with a full 360° / 180° range on them, so they can be shot from any angle.