Oh God is a beautiful problem! I wish I was the author... By the way, another beautiful problem from my favorite proposer... Mr.Golovanov, how did you propose this beauty? Because my solution during the exam is similar to official solution, I posted the official solution. Yes. It is even possible to make the numbers in question differ at most by $1$. Let us color a positive integer with color $r (r = 0, 1, 2)$ if the number of prime factors in the prime factorization of $n$ leaves the remainder $r$ upon division by $3$. In other words, $$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$is colored with the color $\alpha_1 + \alpha_2 + \cdots + \alpha_k$ $(mod\ 3)$. We will check now that this coloring is valid. Let $a_{n,0}$, $a_{n,1}$ and $a_{n,2}$ be the numbers of divisors of $n$ such that their number of prime factors leaves the remainder $0, 1$, and $2$ respectively upon division by $3$. We will prove that the difference of any two of these numbers does not exceed $1$. The proof is by induction on the number of different prime divisors of $n$. If the number of prime divisors is $0$, then $n = 1$, $a_{n,0} = 1$, $a_{n,1} = 0$, $a_{n,2} = 0$. Suppose the statement is true for some $n$, and let $m = np^k$, where $p$ is a prime not dividing $n$. All divisors of $m$ form $k + 1$ groups $G_0, \cdots , G_k$ so that for each $s, 0 \le s \le k$, the group $G_s$ is contains the divisors divisible by $p^s$ and not divisible by $p^{s+1}$. If $s \equiv 0$ $(mod 3)$, the groups $G_s$ contains $a_{n,0}$ divisors of color $0$, $a_{n,1}$ divisors of color $1$ and $a_{n,2}$ of color $2$; If $s \equiv 1\ (mod\ 3)$, then $G_s$ contains $a_{n,2}$, $a_{n,0}$, $a_{n,1}$ divisors of colors $0, 1, 2$ respectively; If $s \equiv 2\ (mod\ 3)$, $a_{n,1}$, $a_{n,2}$ and $a_{n,0}$ divisors of these colors. Clearly a union of the form $G_k \cup G_{k+1} \cup G_{k+2}$ contains equal number of divisors of all three colors. If $s \equiv 2\ (mod\ 3)$, all the groups can be split into such triplets, and $m$ has $5$ equal number of divisors of all three colors. If $s \equiv 0\ (mod\ 3)$, only one group, namely $G_s$, remains outside such triplets, and the differences between numbers of divisors of the three colors are equal to those of $a_{n,0}, a_{n,1}, a_{n,2}$. Lastly, if $s \equiv1\ (mod\ 3)$, the two group outside the triplets are $G_{s-1}$ and $G_s$; they contain $a_{n,1} + a_{n,2}$ divisors of color $0$, $a_{n,2} + a_{n,0}$ divisors of color $1$ and $a_{n,0} + a_{n,1}$ divisors of color $2$. The differences of these numbers also equal to those of $a_{n,0}, a_{n,1}, a_{n,2}$, that is, do not exceed $1$ as desired.

M11100111001Y1R wrote: Oh God is a beautiful problem! I wish I was the author... By the way, another beautiful problem from my favorite proposer... Mr.Golovanov, how did you propose this beauty? Because my solution during the exam is similar to official solution, I posted the official solution.. I'm not sure this is stronger or not ,but I think a similar solution works for every other m instead of 3.

The bound can be improved to difference of $1$. Color $n$ with $\Omega(n)\%3 \in \{0,1,2\}$, which counts the number of prime factors of $n$ with multiplicity. This works by induction on the number of prime divisors of $n$; base case of $1$ divisor trivial. Let $n=p^km$ where $p \nmid m$, and suppose $m$ has $a_0,a_1,a_2$ divisors of colors $0,1,2$ respectively. Adding $3$ to $k$ corresponds to adding $a_0+a_1+a_2$ to the number of divisors of $n$ with a certain color. Thus we only need to check $k \leq 2$ which is easy. $k=0$ gives $a_0,a_1,a_2$ which is immediate, $k=1$ gives $a_0+a_1,a_1+a_2,a_2+a_0$ which can be seen by subtracting these from $a_0+a_1+a_2$, and $k=2$ gives an equal number of all three colors.