Can someone help me with a basic LED set up?
First, if the LEDs really only have 1.45 V forward drop at 100 mA, then they are near IR, not visible.

Second, the best parallel series arrangement depends on what power supplies you have available.

Let's say, for example, that you get a 24 V power supply.  Let's also say that you have a series of LEDs that average to 2.0 V forward drop at their intended 100 mA current.  Ideally, you could string (24 V)/(2.0 V) = 12 LEDs in series.  However, that leaves no room to regulate the current.  At the very least, you'd want 11 LEDs in a string, and a resistor to drop the remaining 2.0 V at the desired LED current.  By Ohm's law, that would be (2.0 V)/(100 mA) = 20 &Omega;.  You could use a 20 &Omega; resistor if you want to live on the edge.  A 22 &Omega; resistor gives you some margin to compensate for the LED forward voltages not being exactly what you expected.  This can easily happen due to temperature and normal part variations, for example.

Once you figure out how many LEDs per string, you then decide how many strings you need.  In the example above, if you used 11 LEDs/string, then you'd need two strings to meet your 15 to 25 LEDs spec.  I'd probably go for 10 LEDs/string with a larger series resistor to keep the current more predictable.  That would still give you 20 LEDs with two strings.

Each string draws 100 mA, so two strings draw 200 mA.  You would therefore need a 24 V supply that could deliver at least 200 mA.  I'd therefore want to see it rated for 250 mA minimum to have some margin.  That would be a 6 W supply.

Added

You now say that these are IR LEDs that really do have about 1.45 V drop at 100 mA.  You should be able to replicate the analysis above using different numbers.

Briefly, (24 V)/(1.45 V) = 16.6, so no more than 15 LEDs per 24 V string.  Your maximum number of LEDs is 25.  Using 24 to keep the strings equal means 12/string.  That's 17.4 V, leaving 6.6 V for the current-setting resistor.  (6.6 V)/(100 mA) = 66 &Omega;, so 68 &Omega; it is.  Note that the resistor will dissipate 640 mW, so it should be "1 W" or more.

For better efficiency, you could use 15 LEDs/string and run them at 80 mA and get the same total light output.  At this point, it should be obvious how to figure out the resistor value and its power dissipation.