How is the linear momentum conserved in this system? Let there be a block on a frictionless surface. Let an agent constantly exert an invariable force, say $F$ on the block and does positive work by displacing the block by $d$ units.
Here, the system is constituted of the agent and the block.
By Newton's third law of motion, the block will exert force $-F$ on the agent and does negative work by displacing the agent by $d$ units. Thus work done on the block $$W = F\cdot{d}$$ and the work done on the agent is $$W' = -F\cdot{d}$$. Thus, the energy of the system is conserved.
Now, since it is an isolated system (agent + block) and no external force acts on the system, the linear momentum of the system must be conserved.
Hmm... Here I got blanked & baffled and couldn't point out how the linear momentum of this system is conserved. The initial velocity of both the components of the system is zero and hence their sum of the momentum is zero initially. Therefore at any moment after that the  their sum of the momentum must be zero and since mass cannot be negative, their velocities must be opposite to each other. But they are moving in the same direction. So how can one's velocity is positive and other's is negative though they are moving in the same direction?
In a word, how is the system's linear momentum conserved? Please help.
[Note: Here sign only denotes direction. Positive velocity means the body is moving in the positive direction & vice-versa.] 
 A: What you described is completely impossible. Your agent must use some sort of engine to generate the force, this engine has to interact with the ground, or the atmosphere, or something else which lies outside your 'block + agent' system. Therefore, your system is not isolated.
A quick example: suppose the agent uses a jet engine. Then the true isolated system is: 'block + agent + jet fuel'. Jet fuel flies backwards with enormous speed, preserving the total linear momentum.
A: 
Let an agent constantly exert an invariable force , say F on the block and does positive work by displacing the block by d unit. Here, the system is constituted of the agent and the block. By Newton's third law of motion, the block will exert force -F on the agent and does negative work by displacing the agent by d unit. 

This is not physically possible for an isolated agent+block system. This setup is the horizontal equivalent of lifting oneself by ones bootstraps.
Suppose the agent and block have masses M and m, respectively. The easiest way to see what will happen is to analyze things from the perspective of a frame in which the agent and the block are initially at rest. Assume the agent exerts some force F on the block in the +x direction, for some time t, moving the block through some distance d in the +x direction. If the force is constant, the distance d is given by $d=\frac 1 2 \frac F m t^2$. Since the block moves in line with the force, the work on the block is $Fd = \frac 1 2 \frac {F^2} m t^2$. The block's velocity after applying this force is $v=\frac F m t$. The change in the block's kinetic energy is $\frac 1 2 m v^2 = \frac 1 2 \frac {F^2} m t^2$, which is equal to the work done on the block.
What about the agent? The block exerts an equal but opposite force on the agent. This force pushes the agent in the -x direction. The agent travels a distance $D=d \frac m M$ during the time $t$ during which the force is applied, but in the -x direction. The work on the agent is positive as the agent moves with rather than against the force applied by the block: $FD = \frac 1 2 \frac{F^2} M t^2$. The agent's velocity after applying this force is $V=- \frac F M t = -\frac m M v$. The change in the agent's kinetic energy is $\frac 1 2 M V^2 = \frac 1 2 \frac {F^2} M t^2$, once again equal to the work done on the agent. The work done on the agent and on the block is $W_{\text{tot}} = F (d+D) = Fd(1+\frac m M)$. This is of course equal to the total change in kinetic energy.
What about linear momentum? The linear momentum after the interaction is complete is $mv + MV = mv - M\frac m M v = 0$. Linear momentum is conserved.
Note that kinetic energy is not conserved. There is no such thing as conservation of kinetic energy, or conservation of work. The concept is conservation of energy. Some form of energy was needed to generate that push. For example, that push might have come from releasing a compressed spring or using a human arm. Releasing the spring or pushing with an arm converts potential energy into kinetic energy.
