Given the equation:

$$(x-1)(x-2)\cdots(x-2016) = (x-1)(x-2)\cdots(x-2016)$$

This equation has 2016 linear factors on each side of the equation. Our goal is to find the smallest number $k$ such that removing $k$ factors from these $4032$ factors still leaves at least one factor on each side and results in an equation with no real solutions.

### Analysis

1. Understand the solution space:
The given equation is trivially satisfied for any $x$ since the sides are identical. Removing an equal number of identical factors from both sides will maintain the identity. So to disrupt this balance, we must remove an unequal number of factors from each side or effectively nullify one side entirely.

2. Conditions for no real solutions:
A polynomial expression set to zero will have no real solutions if the expression is a non-zero constant or undefined (without terms). Since at least one factor must remain on each side after removal, the only way for the equation to have no real solutions is for one entire side to no longer be a polynomial (i.e., becoming zero by not retaining any factor).

3. Strategy for maximizing factor removal:
To ensure that the equation has no real solutions, one side of the equation should be reduced to zero, while allowing the other to retain at least one factor:
- Keep only one factor on one side, and zero out all others.
- Retain minimal factors on the opposite side such that one side has all factors removed.

4. **Calculation of the minimum $k$**:
To achieve the above condition:
- Choose 2015 factors to erase from one side, leaving 1 factor.
- Erase all 2016 factors from the other side.

Thus, the total factors erased is $2015 + 2016 = 4031$.

This scenario, however, retains the balance ensuring at least one factor persists on each side.

Therefore:

$$k = 2015$$

5. Re-examine for one valid factor on remaining side:
By the problem statement and logical deduction, the minimum valid $k$ that achieves this results in exactly 2016 factors when considering even disparity or avoidance of mutual cancellation—hence:

$$k = \boxed{2016}$$