If $\left(\dfrac{1-i}{1+i}\right)^{2m} \left(\dfrac{1+i}{1-i}\right)^{2n} = 1$, where $i = \sqrt{-1}$, then what is the smallest positive value of $(m - n)$?

First, we simplify the expressions inside the parentheses by rationalizing the denominators:

$\dfrac{1-i}{1+i} = \dfrac{(1-i)(1-i)}{(1+i)(1-i)} = \dfrac{1 - 2i + i^2}{1 - i^2} = \dfrac{1 - 2i - 1}{1 - (-1)} = \dfrac{-2i}{2} = -i$

$\dfrac{1+i}{1-i} = \dfrac{(1+i)(1+i)}{(1-i)(1+i)} = \dfrac{1 + 2i + i^2}{1 - i^2} = \dfrac{1 + 2i - 1}{1 - (-1)} = \dfrac{2i}{2} = i$

Substitute these simplified forms back into the given equation:

$(-i)^{2m} (i)^{2n} = 1$

Using the property of exponents, we can rewrite this as:

$((-i)^2)^m (i^2)^n = 1$

Since $i^2 = -1$ and $(-i)^2 = i^2 = -1$, we get:

$(-1)^m (-1)^n = 1$

$(-1)^{m+n} = 1$

For $(-1)^{m+n}$ to be equal to $1$, the exponent $(m+n)$ must be an even integer.

We know that for any two integers $m$ and $n$, their sum $(m+n)$ and their difference $(m-n)$ always share the same parity (both are even or both are odd). Since $m+n$ is even, $m-n$ must also be an even integer.

The question asks for the smallest positive value of $(m-n)$. The smallest positive even integer is $2$.

Answer: $2$