Let $k$ be a fixed even positive integer, and let $N$ be the product of $k$ distinct primes $p_1, p_2, \ldots, p_k$. Let $a$ and $b$ be two positive integers such that $a, b \leq N$. Define the sets:
$$S_1 = \{ d \mid d \mid N, a \leq d \leq b, \text{ and } d \text{ has an even number of prime factors} \},$$
$$S_2 = \{ d \mid d \mid N, a \leq d \leq b, \text{ and } d \text{ has an odd number of prime factors} \}.$$

We aim to prove that:
$$|S_1| - |S_2| \leq \binom{k}{k/2}.$$

Consider the factors of $N$ as sets of the primes. Each factor $d$ of $N$ can be represented by a subset of the set of primes $\{p_1, p_2, \ldots, p_k\}$. Let $s_1, s_2, \ldots, s_{2^k}$ be these subsets, and let $f_1, f_2, \ldots, f_{2^k}$ be the corresponding factors.

By Sperner's Theorem and Dilworth's Theorem, we can cover the poset of these subsets with $\binom{k}{k/2}$ chains. For each chain $C$, let $C_1$ and $C_2$ be the sets of subsets that have even and odd sizes, respectively.

For any chain $C$, the difference $|C_1| - |C_2| \leq 1$. This is because if $a \leq f_i \leq b$ and $s_i \subset s_k \subset s_j$, then $a \leq f_k \leq b$, which is a trivial observation.

Summing over all $\binom{k}{k/2}$ chains, we get the desired result:
$$|S_1| - |S_2| \leq \binom{k}{k/2}.$$

The answer is: $\boxed{\binom{k}{k/2}}$.