Hi Zacky, the answer is yes because of the losses in the coil.
  
Although you are not changing the energy in the magnetic field  (as you correctly assume in your first paragraph) or in the spring, you are having to supply the heat losses due to coil resistance.  All you can do is reduce the coil current to a level that just holds the contacts closed.

With resistance neglected, there is no energy required once the coil is pulled in.

On the electrical side, voltage would equal zero in the theoretical steady state dc system with all resistance =0. electrical power = voltage times current = 0 times current = 0.

On the mechanical side:
mechanical power = force times velocity = force times 0 = 0
since velocity = 0 once the coil is pulled in.
velocity = 0 once the coil is pulled in so

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That is fine, once the contacts closed , there will be no
energy cosumption after that ( if the coil resistace is neglected}.
But at the same time we have to pass enough  current through the coil to get the magnetic field required to produce pulling force equal to or greater than the opposing
spring force.

Now, if we know the value of the force of the the spring
When the cotacts are fully closed, whow we can calculate
the minimum current required to keep the contacts closed ?

The relay s basically a solenoid that pulls the contacts against the spring. You need to calculate/design the solenoid such that at the current that you can provide the force will enough to overcome the spring force and provide enough contact force too.

http://www.webspawner.com/users/israelkk/

You can do some back of the napkin magnet calculations to know how much force the relay will produce but because of non-linearities & fringing in the airgaps, etc., it will not be very accurate.  For a better estimate, you can download FEMM for free. http://femm.foster-miller.net/

Ukpete, what about AC relay, does it need to reduce the coil current after full pull up? - or the magnetic circuit permeabilty will go up and the coil inductace will increase and the current reduces automaticaly ?
Thanks

zacky, it is true that the ac relay current will decrease slightly when the relay closes, though I suspect that the resistance will dominate the overall impedance.  The other problem with the ac relay is that the force decreases during the zero-crossover part of the ac cycle, so you may need to make the current higher to compensate.

AC electromagnet roughly needs twice the current than a DC electromagnt.

zacky, you may be able to do an approximate calculation of the current required to hold the solenoid against the spring.  As given in other recent threads, the force between two poles is given by:
F = B²A / 2μ₀
(all in SI units; A is the pole face area, B is flux density, μ₀ is permeability of free space)
Strictly speaking this is the force between two facing poles separated by an airgap, applicable provided the airgap isn't large compared with the pole face diameter or width (otherwise you will have difficulty deciding what value of B to use).  

I'm not sure what your magnetic circuit looks like; if when your "contacts" close the airgap in your magnetic circuit also closes, then just assume a very small airgap still exists (experiment by decreasing the value of airgap until you arrive at a maximum value of B) and calculate the total reluctance of the magnetic circuit.  Then find how many ampere-turns you need to match the spring force when the contacts are closed.

This discussion really misses the point. The energy in the magnetic field of the relay coil is negligible compared to the power used to keep a relay energized. After the first fraction of a second the energy efficiency of a relay will approach zero. One method to improve on this is a mechanically latching relay, but this is seldom justified due to cost and complexity. It can be compared to lifting a bucket of water and holding it. The energy used to hold it up by hand does zero work but after a few seconds you will expend as much energy holding it as you did lifting it.