When we write that $F = -\nabla V$ , what would happen if we ommit the (-) minus sign I have had this question for a long time. In classical mechanics, if we choose $\mathbf F = -\mathbf \nabla \, V,$ with the minus sign, we can proof the work - kinetic energy theorem. What are the consequences of omitting that minus sign?
 A: Take a simple 1D example. We'll plot gravitational potential energy as a function of horizontal position for the side of a hill:

We normally define $F = -\nabla V$ and that means the force points downhill i.e. if we let the particle go it will roll downhill. If instead we defined $F = \nabla V$ the force would point uphill i.e. if we let the particle go it will roll uphill.
Rolling downhill is fine because the decrease in potential energy is accompanied by an increase in kinetic energy, so total energy is conserved. If the particle rolls uphill that means both its kinetic and potential energy are increasing so conservation of energy is violated.
A: One way of looking at the change is that you are changing the definition of potential from

The potential at a point is the work done on a unit mass in going
  from an arbitrarily chosen zero of potential to the point

to

The potential at a point is the work done by a unit mass in going
  from an arbitrarily chosen zero of potential to the point

So in the first definition you do work on the mass and in the second definition the mass does work on you.
A: The gradient of a scalar function always points in the direction of greatest ascent. That is, a scalar function has the most rapid increase in that direction. If you imagine walking around a mountain, the gradient of the height function of the mountain at each point will point towards the steepest direction.
On the other hand, forces in a conservative field seek to minimize potential energy. Thus, they move in the direction of greatest descent, which is in the opposite direction of the gradient, hence the minus sign.
A: The gradient of a function $\nabla V$ is as you know a vector and this vector points to the (locally) steepest ascent. i.e. if you are on the hill the vector $\nabla V$ points to the tip of the hill and when you are on top of the hill the vector vanishes because there is no "steepest ascent" at that point. You can prove this fact mathematically but I'll omit the derivation.
If we had $F = \nabla V$ then we would introduce a minus sign in the potential term. e.g. instead of the Coloumb potential $V= a\, q Q /r$ we would get $V = - a\, q Q/r$, where $a$ is a constant to make everything sensible and the equation would in both cases read $F = a \, q Q/r^2$. Mathematically there is no difference in both cases. There is however an intuitive one. Look at the plots of the potentials for which both of the charges is negative or positive.

In the first case you can imagine a ball putting on the blue curve in a gravitational field and just roll down the slope, which represents the actual situation really good but in the second case you don't have that intuition anymore. Since the ball actually go uphill because that is what $\nabla$ suppose to do. It shows in the direction of the steepest ascent. 
Long story short the minus sign is barely a convention to make our lives easier and make physics more intuitive. 
A: There are no consequences. It's a convention that negative is for attraction forces and positive is for repulsion. This leads to forces tending to reduce the potential when attracting. You remove that sign, then attraction forces because positive, and forces will tend to increase the potential. The worst that would happen is a few inconsistencies within whatever you're calculating, because we accept that convention in all physics.
You have to wrap your mind around the idea that we're modeling nature, meaning that we give meaning to these equations, signs and numbers. Nothing will happen if you switch the sign unless you give it a meaning that will make a difference.
Note: This is all classical mechanics. Modern physics is not 100% symmetric within such quantities.
A: This the same as changing $V$ to $-V$. For example, if $V$ described a potential well then changing it to $-V$ would describe a bump.
