Electric force and potential difference 
When an electron moves from lower potential to higher potential and the work done by electric force is positive.

Therefore we will conclude that whenever a negative charge moves from a lower potential to higher potential, work done by electric force is positive and when a positive charge moves from lower potential to higher potential the work done by electric force is negative. Why? What is the concept?
 A: The increment of work done by a force is $dW = {\bf F} \cdot d\bf x$.
The electric force on a charge is ${\bf F} = q {\bf E}$, where the electric field is ${\bf E} = -\nabla \phi$.
That means that the increment of work done on a charge by the electric force is $dW = -q\nabla \phi \cdot d{\bf x} = -q d\phi$.
So, by moving from a lower potential to a higher potential $d \phi$ is positive, and $-q d\phi$ is positive for a negative charge and negative for a positive charge.
A: Say, you've got a negative charge and a plate near it. Now suppose that the plate has a positive charge. So, you can say that the "location" of the positively charged plate is at a higher potential than the "location" of the negative charge. 
Also, it is obvious that the positive plate will attract the electron towards it, and the electron would move from lower potential to higher potential i.e. the displacement is along the direction of electric force, and work done is positive
Now, in case you pull the electron farther away from the positive plate, there is going to be a decrease in its potential. The electric force is in the same direction (towards the positive plate) while the displacement is opposite and hence the work done is negative
A: 
Therefore we will conclude that whenever a negative charge moves from
  a lower potential to higher potential, work done by electric force is
  positive

That is correct. The electron gains electrical potential but loses electrical potential energy. The energy for doing positive work on the electron comes from the electrical potential energy of the charge field system. If the electron is unrestrained it will accelerate and gain kinetic energy equal to the loss of electrical potential energy. 

...and when a positive charge moves from lower potential to higher
  potential the work done by electric force is negative. Why? What is
  the concept?

Correct, but you are missing an important additional point. 
An external (to the field) agent is needed exert a force to move a positive charge from lower potential to higher potential against the repulsive force of the electric field. Since the direction of the force exerted by the external agent is the same as the displacement of the charge, it does positive work. If the charge starts and ends at rest, the electric field does an equal amount of negative work since its force is in the opposite direction of the displacement. Assuming there are no dissipative forces involved, the net result is the negative work done by the field takes the energy supplied to the positive charge by the external agent and stores it as electrical potential energy. 
The gravity analogy, in the absence of dissipative forces, is you do positive work lifting an object to rest at a height $h$  above the ground and gravity does an equal amount of negative work taking the energy you supplied and storing it as gravitational potential energy of $mgh$.
Hope this  helps.
