HMMT November, 2021

Solution:
I claim the answer is $120$. First, note that $120 = 2^{3} \cdot 3 \cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5) = 15 \cdot 4 \cdot 6 = 360$. Thus, we need to split the divisors into groups summing to $120$. But then we can just take $\{120\}$, $\{20, 40, 60\}$, $\{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 24, 30\}$. Thus, $120$ works.

Now we need to show $120$ is the lowest. Let $s(n)$ be the sum of divisors. Since $n$ will be in one of the piles, we need $s(n) \geq 3n$. First, we claim that $n$ must have at least $3$ distinct prime divisors. Surely, if it had $2$ distinct prime divisors, say $p$ and $q$, so that $n = p^{a} q^{b}$, then the sum of divisors is
$$\left(1 + p + p^{2} + \ldots + p^{a}\right)\left(1 + q + q^{2} + \ldots + q^{b}\right) = p^{a} q^{b}\left(1 + \frac{1}{p} + \ldots + \frac{1}{p^{a}}\right)\left(1 + \frac{1}{q} + \ldots + \frac{1}{q^{b}}\right)$$
However, the expression $1 + \frac{1}{p} + \ldots + \frac{1}{p^{a}}$ is maximized when $p$ is minimized, and further, as $a$ is finite, must be at most $\frac{1}{1 - \frac{1}{p}} = \frac{p}{p-1}$. Thus, the sum of divisors is less than
$$p^{a} q^{b} \frac{p}{p-1} \frac{q}{q-1} \leq n \cdot 2 \cdot \frac{3}{2} = 3n$$
Thus, $n$ can't have $2$ distinct prime divisors and must have at least $3$ distinct prime divisors.

As we already discovered $120$ works, we need not worry about $4$ distinct prime divisors, as the value of $n$ would be at least $2 \cdot 3 \cdot 5 \cdot 7 = 210$. We now work through the numbers with $3$ distinct divisors. If $2$ is not one of them, then the only number that works is $105 = 3 \cdot 5 \cdot 7$, which has a sum of divisors that is not large enough. Therefore, $2$ must be a prime divisor of $n$. Additionally, if $3$ is not a divisor, then our options are $2 \cdot 5 \cdot 7$ and $2 \cdot 5 \cdot 11$, which also do not work. Therefore, $3$ must also be a prime divisor. Then, if $5$ is not a prime divisor, then if $n$ is $2 \cdot 3 \cdot p$, it has a sum of divisors of $(1+2)(1+3)(1+p) = n \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{p+1}{p}$, which is only at least $3n$ if $p$ is exactly $2$, which is not feasible. Additionally, if we use $2^{2}$, then the sum of divisors is $(1+2+4)(1+3)(1+p) = n \cdot \frac{7}{4} \cdot \frac{4}{3} \cdot \frac{p}{p+1}$, so $\frac{p+1}{p} > \frac{9}{7} \Longrightarrow p < 4.5$, which also can't happen. Further, we can't have $3^{2}$ be a divisor of $n$ as $2 \cdot 3^{2} \cdot 5$ is the only value less than $120$ with this, and that also does not work. Lastly, we just need to check $2^{3} \cdot 3 \cdot p$, which has a sum of divisors of $(1+2+4+8)(1+3)(1+p) = n \cdot \frac{15}{8} \cdot \frac{4}{3} \cdot \frac{p+1}{p} = n \cdot \frac{5}{2} \cdot \frac{p}{p+1}$, so $p=5$ and that works. This means that $n=120$ is the smallest value for which $s(n) \geq 3n$, and thus is our answer.