If the person is moving the block in such a way so that the sum of the
forces acting on it is equal to zero, how can he be moving it at all?
Consider a person pushing the block of wood along a surface with friction where the force due to friction (a force proportional to the speed of the block) exactly cancels the pushing force from the person.
The forces add to zero so the block does not accelerate. However, in order for the forces to add to zero, the block must be moving.
This addendum addresses the (latest) edited version of the question:
The first gets the job done in one-half the time that the second
takes. Did one of the workers do more work than the other?
First let's ignore the accelerations at the beginning and end.
Work is force through distance. A brick lifted with constant speed against the pull of gravity to a given height requires a certain amount of work to be done by the worker regardless of the time spent lifting.
So, comparing the amount of work done while the bricks move with constant speed, there is no difference.
However, there is a difference in the power since power is the rate at which work is done.
If the same amount of work is done in half the time, the associated power is double. One worker does work at twice the rate of the other.
Now, let's look at the accelerations. To start a brick moving requires that the worker do work on the brick such that the brick gains kinetic energy.
But to stop the brick moving requires that the brick do work on the worker such that the brick loses that kinetic energy.
Thus, the work associated with the accelerations (ideally) cancel and don't factor into this calculation.
Let's tie this together with your original question:
How can an object with zero acceleration move?
Clearly, in this problem, each brick is at rest, is then briefly accelerated to some speed, moves at this speed for some distance, and then is briefly decelerated to rest.
During the portion in which a brick moves upward with constant speed, the worker provides a force which cancels the force of gravity. The momentum of the brick is constant since there is zero net force acting on the brick.
But, the brick is moving upward due to the brief acceleration where it gained momentum and kinetic energy.
Because it's not any work, but the work done by a force that produces a displacement.
In the scenario you describe, somehow that force is not doing any work on the particle. This could be because the particle is restricted by another force to not go perpendicular and then the sum of forces in the perpendicular direction is zero.
In the second scenario, with the box and the normal force, it's the same. That force doesn't do any work since in the direction of that force there is zero movement. Which is analogous to say that the cosine of the angle between the displacement and such force is 90°.
Best Answer
In introductory problems about work you're normally taught that it's force times distance:
$$ W = F \times x $$
and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is:
$$ W = \int^{x_f}_{x_i} F(x) dx $$
The force $F(x)$ can be a function of $x$, and to get the work we integrate this force from the starting point $x_i$ to the final point $x_f$. Because $F(x)$ can vary we can make $F > mg$ at the beginning to accelerate the box then make $F < mg$ towards the end so the box slows to a halt again.
DavePhD comments that work is not a state function, and in general this is true. However in this case the work done is equal to the change in potential energy so as long as the box starts at $x_i$ at rest and ends at $x_f$ at rest we'll get the same work done regardless of the exact form of $F(x)$.
If you're really determined to have $F$ constant then start with $F > mg$ at the beginning and $F < mg$ at the end, then gradually reduce the initial value of $F$ and increase the final value to make the force more constant. This will cause the time taken to move the box from $x_i$ to $x_f$ to increase. The limit of this process is a completely constant value for $F$, in which case it takes an infinite time to move the box.