You have a 24v 400VA transformer ... or 400VA/v = 16.6 A.
If you convert this to DC you get a peak voltage (before voltage drop on diodes on rectifier) of 24 x 1.414 = 33.9v and your current drops to about 0.68 x 16.6 = 11.2A.
If you convert this to DC and use a large inductor after the bridge rectifer, you get a peak DC voltage of about 0.9 x 24 = ~ 22v but your current remains at around 0.94 x 16.6 = 15.6A . Honestly, I have no idea how large the inductor would have to be for such high current.
Going back to the first choice (just bridge rectifier and capacitors) like I said, you'll have a peak voltage of about 34v before the voltage drop on diodes and capacitors.
You have two diodes always working in a bridge rectifier, and a bridge rectifier rated for such high currents would have about 0.8-1.1v drop per diode, so you're losing about 2v on the bridge rectifier. Therefore, you'll have a peak voltage of about 32v before the capacitors.
Ideally though, you'll put two bridge rectifiers in parallel to reduce the power dissipated in each bridge rectifier, because bridge rectifiers are derated at higher temperatures.
For example, look at this 35A bridge rectifier :
http://www.digikey.com/product-detail/en/GBJ3510-BP/GBJ3510-BPMS-ND/3191597 If you check the datasheet, you can see that it's rated for 35A at up to 100c but it can only do about 7A without a heatsink. It has a voltage drop of about 0.8v at 2A , 1v at 10A per diode..
So you know that at 11A it will dissipate 2v x 11 = ~ 22 watts and therefore you need to keep it under 100c with a decent heatsink.
If you put two in parallel, they'll each dissipate half that current amount, more or less and you're getting almost to the point where they could work without a heatsink, so you won't need a large heatsink or you won't have to worry much about it.
Anyway, going back to the subject... you now have a peak of 32v DC after the diodes and their voltage drop and about 11A. Now you need to figure out how much capacitance to use.
The capacitance affects how much the voltage will sag at a particular current used. A "close enough" formula is :
C = Current / (2 x FrequencyAC x Vripple) where Frequency AC is the mains frequency (60 Hz in US and some countries, 50 Hz in Europe) and Vripple is how much you're willing to let the voltage go down.
For 11A and 2v ripple and EU frequency, we have C = 11 / (2 x 50 x 2) = 11 / 200 = 0.055 Farads or 55000 uF - so with your 50000 uF capacitors, you'll have a voltage hovering around 29.5 and 32v when you use 11A. At lower power consumption, the voltage will be closer to 32v more often.
In practice, 55000 uF is A LOT. Normally, it makes little sense to use more than 10000-220000 uF and just accept that the voltage may go down a bit more - a switching regulator will just adjust it's pwm to maintain the output voltage where you want it. It's also usually more expensive to buy such large capacitors and it takes more space, compared to just using a switching regulator or mosfets with better efficiency, or a slightly larger power transformer.
So now you have to find a switching regulator that would accept.. let's say 25v - 35v and capable of producing 17v 15-20 amps ... I'm not quite sure you'd be able to do 20 amps, because 17x20 amps is 340 watts and at best you'd have about 30v x 11 = 330 watts and then the switching regulator will be about 90% efficient, so you lose another 20-30 watts, but about 17v @ 17-18a should be doable. 260w like you said in your last post should work.
My advice from this point forward would be to go on ti.com and create an account and then use their Webbench software to find something suitable with those parameters (25-35v in, 17v @ 20a out) and see what circuit would work best for you. The software gives you mosfet suggestions (ex pick through hole mosfets instead of surface mount stuff), inductor choices etc etc
For example, you see below two results generated by webbench, that do 28-32v in, 17v @ 18a out, with LM5119 or LM25119, and they both use pretty much the same parts ( four mosfets, some inductors, lots of resistors and capacitors and diodes) :
later edit: and if you can go down to about 140w output (17v @ 10A) you can simply the circuit and use just one mosfet and an easier to solder part, LM5088.. see circuit3.png