The transformer is rated at 24 V at full current, when the load is resistive. The peak voltage is only at 1.4142 times that for a really short time. In fact, the voltage is only more than 30 V for around 1/3 of the time, so if you try to take all your power only when the voltage is more than 30 V, you will get much more voltage drop and the peak voltage will be reduced.
Of course, a lot of modern electronic appliances work like that, and take most of their current at the voltage peak. Almost any appliance that runs on DC internally will take little current unless the voltage is near the peak, although there are now some rules to make larger electronic appliances do that to a lesser extent.
The result is that the mains voltage already has the peak voltage reduced, so the 24 V output will have a lower peak than you would calculate. You then need to subtract:-
Diode drop
LM317 minimum voltage difference
LM317 reference voltage
LED tolerances
and the one that seems to have been forgotten, capacitor ripple voltage.
You said you are using 1000 μF, and are trying to run 9 strings at 22 mA so 198 mA. If the mains is at 50 Hz (which I am assuming as it is 220 V) then the capacitor has to support the LEDs for around half a cycle or 10 ms. That means that the charge is 0.00198 C, so the capacitor which has a capacitance of 0.001 F will reduce by 1.98 V during that time
If we add that lot up, we get:-
30 V peak (assuming we are going to be happy with current only flowing 1/3rd of the time)
1.6 V drop across 2 diodes
2 V minimum LM317 difference
1.25 V LM317 reference
1 V LED tolerance (10 at 0.1 V each, just a guess)
2 V capacitor ripple
which comes to 22.15 V.
So it is no surprise that your 26 - 27 V string isn't working as you want it to.
Now I haven't measured anything before posting this, apart from noticing that the mains is tending to be clipped. You might well find that a few of these items aren't as bad as I am suggesting. However, you need to do calculations like this to get a circuit to work in the real world.
There is also the supply voltage tolerance. Mains in Europe is generally 230 V ±6%, so can be 216 V, so that takes another 2% off the 30 V for worst-case conditions.
You can reduce the diode drop by using Schottky diodes. A larger capacitor will give less ripple. There are other current regulator circuits that have lower voltage drops.
Also, a 2 x 12 V, 1 A transformer would be expected to run at around 13.8 V with no load (
https://www.farnell.com/datasheets/1829278.pdf) so I would expect the open circuit voltage to be around 38 V, as the diode drop will only be around 1 V with no load. If you are getting 33 V at no load, the transformer voltage will almost certainly be quite a bit lower than 12 V per winding when loaded, and the transformer is not meeting specification.