, 2019

Solution:

First, we show that $N$ cannot have three distinct prime divisors. For the sake of contradiction, suppose $p q r \mid N$ for three distinct primes $p, q, r$. Then by the problem statement, $(p, q, 1), (p, r, 1)$, and $(q, r, 1)$ have three distinct colors, so $(p, q, r, 1)$ has four distinct colors. In addition, $(p q, r, 1), (p q, p r, p)$, and $(p q, q r, q)$ have three distinct colors, so $(p q, p, q, r, 1)$ has five distinct colors, contradicting the fact that there are only four possible colors.

Similarly, if $p^{3} q \mid N$ for some distinct primes $p$ and $q$, then $(p, q, 1), (p^{2}, q, 1), (p^{3}, q, 1), (p^{2}, p q, p), (p^{3}, p q, p)$, and $(p^{3}, p^{2} q, p^{2})$ are all triples with distinct colors, so $(1, q, p, p^{2}, p^{3})$ must have five distinct colors, which is again a contradiction. In addition, if $p^{2} q^{2} \mid N$ for some distinct primes $p$ and $q$, then $(p, q, 1), (p^{2}, q^{2}, 1), (p^{2}, q, 1)$, and $(p, q^{2}, 1)$ are all triples with pairwise distinct colors, so $(1, p, q, p^{2}, q^{2})$ must have five distinct colors, another contradiction.

We are therefore left with two possibilities:

- Case 1: $N = p q$

In this case, the only triple of factors that must have pairwise distinct colors is $(p, q, 1)$. We have $4 \cdot 3 \cdot 2 = 24$ choices for these three, and $4$ choices for $p q$ itself, giving $4 \cdot 24 = 96$ multichromatic colorings.

- Case 2: $N = p^{2} q$

In this case, the triples of pairwise distinctly colored factors are $(p, q, 1), (p^{2}, q, 1)$, and $(p^{2}, p q, p)$. From this, we see that $(1, p, q, p^{2})$ must have four distinct colors, and the color of $p q$ must be distinct from $p$ and $p^{2}$. There are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways to assign the four distinct colors, $2$ ways to assign the color of $p q$ after that, and $4$ ways to color $p^{2} q$ after that, giving a total of $24 \cdot 2 \cdot 4 = 192$ monochromatic colorings.

Therefore, there can be at most $192$ multichromatic colorings.