Given the problem with positive integers $m$ and $n$ such that $m \leq 2000$, and $k = 3 - \frac{m}{n}$. We are tasked to find the smallest positive value of $k$.

Firstly, to ensure $k$ is positive, we need:
$$k = 3 - \frac{m}{n} > 0,$$
which implies:
$$3 > \frac{m}{n}.$$

Rearranging gives:
$$n > \frac{m}{3}.$$

Since $m$ and $n$ are integers, $n$ must be greater than $\frac{m}{3}$, so we set:
$$n \geq \left\lceil \frac{m}{3} \right\rceil.$$

The expression for $k$ becomes:
$$k = \frac{3n - m}{n}.$$

To minimize the positive value of $k$, we need the smallest possible $n$ that satisfies the condition. Setting the equality $n = \left\lceil \frac{m}{3} \right\rceil$ leads to:
$$n = \left\lceil \frac{m}{3} \right\rceil.$$

Substitute $n = \left\lceil \frac{m}{3} \right\rceil$ into the expression for $k$:
$$k = \frac{3\left\lceil \frac{m}{3} \right\rceil - m}{\left\lceil \frac{m}{3} \right\rceil}.$$

To find the smallest positive $k$, consider the smallest value for which this fraction can exist. For $\left\lceil \frac{m}{3} \right\rceil$ to be as close as possible to $\frac{m}{3}$, let $m = 3t + 1$ or $m = 3t + 2$ for the smallest integer change.

By setting $m = 2000$, we have:
$$n = \left\lceil \frac{2000}{3} \right\rceil = 667.$$

This gives:
$$k = \frac{3 \times 667 - 2000}{667} = \frac{2001 - 2000}{667} = \frac{1}{667}.$$

Thus, the smallest positive value of $k$ is:
$$\boxed{\frac{1}{667}}.$$