Help making a 205 LED lamp
My recommendation is that you go for a safe voltage as this effectively eliminates electric shock hazard concerns. 24 V would be a good choice and power supplies are plentiful.

You can now sort out your LEDs by type and group each type in a series string to make up about "20 V worth" of LEDs and calculate a series resistor to drop the remaining 4 V on each string. \$ R = \frac {V}{I} = \frac {4}{20m} = 200 \ \Omega \$. I've chosen 4 V as a balance between having enough resistance to control the current adequately versus wasting power as heat in the resistors. You can adjust to suit the maths but don't go too low.

Then parallel each string and connect to your 24 V supply.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Series-parallel arrangement for 24 V supply.




  If I were to work with 22,5 V worth of LEDs (so 1,5 V remaining on each string) it would amount to a 75 ohm resistor, correct?


Ideally, for driving LEDs you want a current source. An ideal current source has an infinite output impedance - i.e., a very high series resistance. The lower you make your series resistor the poorer a current source it makes. Let's do the calculations for your 9 LEDs, 22.5 V, with a 75 Ω series resistor per string and work out what happens if the headroom changes by 1 V either due to a variation in the LEDs VF or a variation in the PSU output voltage.


Nominal headroom: 1.5 V. \$ I = \frac {V}{R} = \frac {1.5}{75} = 20\ \mathrm{mA} \$.
1 V low, 0.5 V. \$ I = \frac {V}{R} = \frac {0.5}{75} = 6.6\ \mathrm{mA} \$ (-66%).
1 V high, 2.5 V. \$ I = \frac {V}{R} = \frac {2.5}{75} = 33\ \mathrm{mA} \$ (+66%).


Now let's reduce your string of nine LEDs to eight with a nominal voltage of 20 V and a series resistor of 200 Ω and repeat the calculations.


Nominal headroom: 4 V. \$ I = \frac {V}{R} = \frac {4}{200} = 20\ \mathrm{mA} \$.
1 V low, 3 V. \$ I = \frac {V}{R} = \frac {3}{200} = 15\ \mathrm{mA} \$ (-25%).
1 V high, 5 V. \$ I = \frac {V}{R} = \frac {5}{200} = 25\ \mathrm{mA} \$ (+25%).


You should be able to see from all this that the 200 Ω / 8 LED solution will regulate the current better if there is variation of the LED VF due to manufacture or temperature or variation in the supply voltage. This is the background to my recommendation in the original answer.