By simple calculations, the 0.5 L is about 500 g (Assuming the density of water = 1 g/ml). So you can dissove 5 times the quantities that you have mentioned in your question:

1- 36x5 = 180 g NaCl, since you have only 140 g NaCl, all NaCl will dissolve.(If it is alone)

2- 37x5 = 185 g KCl, so all KCl will dissolve also. .(If it is alone)

3- 56 x 5 = 280 g MgCl2, so all MgCl2 will dissove also..(If it is alone)

4- 100 x 5 = 500 g CaCl2, so all CaCl2 will dissove also. .(If it is alone)

But if all of them togother, the most one that is soluble it dissolves first.

So,

1- All 20 g CaCl2 will dissolve completely and this represents (20/500) x 100% =4.00% CaCl2 of the total solubility

2- (270/280) x 100% = 96.43%.MgCl2 of the total solubility.

3- If we add them together 4.00 + 96.43 we will get 100.43%, so there will be some of the MgCl2 will not dissolve which is equal to 0.43% x 280 = 1.20 g

3- So all CaCl2 will dissolve and 268.80 g of MgCl2 will dissolve.

4- All the quantity of KCl and NaCl will precipitate.

Thank you for your complete answer

Is it possible that these salts are dissolved simultaneously and in proportion to their solubility in water, or the salt that is more solubilized, it is first dissolved, and then the next salt is solved ?!

How much should be the solubility of salts in water, so that they can not be solved simultaneously?

I suggest this work:

Article Thermodynamic modeling of KCl + H2O and KCl + NaCl + H2O sys...