HMMT February

Solution:
We work mod $2025$, so the condition becomes that for any $a$, $b \in S$, we have $\frac{a + b}{2} \in S$.

First, we prove that $S$ must be an arithmetic sequence. Observe that if $S$ is balanced, then so is the shift $S + k = \{s + k \mid s \in S\}$ for all $k$, so we can assume $0 \in S$. Let $s$ be an element of $S$ such that $d = \gcd (s, 2025)$ is minimal. Observe that for any $t \in S$, we have $\frac{t}{2} = \frac{0 + t}{2} \in S$. Thus, $\frac{s}{2^n}$ is in $S$ for all $n$. Because $\gcd (2, 2025) = 1$, this implies $2^n s \bmod 2025 \in S$ for all $n$. We prove the following claim.

Claim 1. For all positive integers $m$, we have $ms \in S$.

Proof. We proceed with induction on the number of 1's in the binary representation of $m$.

Base Case: $m$ has one 1 in binary, so $m$ is a power of 2. Then $ms \in S$ as noted above.

Induction Step: Assume the claim holds for all $m$ which have $k$ 1's in their binary representations. Suppose $m$ has $k + 1$ 1's in its binary representation. Let $2^{n}$ be the largest power of 2 which is at most $m$. Then $m - 2^{n}$ has $k$ 1's in its binary representation, so $2(m - 2^{n})$ does as well. By the induction hypothesis, $2(m - 2^{n})s \in S$. Also, $2^{n + 1}s \in S$, so
$$m s = \frac{2^{n + 1}s + 2(m - 2^{n})s}{2} \in S,$$
as desired. $\square$

It follows that every multiple of $s$ is in $S$. The multiples of $s$ are precisely the multiples of $d$, so $S$ contains every multiple of $d$. Now assume for sake of contradiction that $S$ contains some element $t$ which is not a multiple of $d$, so we can write $t = cd + r$ such that $0 < r < d$. Then $2t \in S$, so
$$r = \frac{2t + (- 2c)d}{2} \in S.$$
But then $\gcd (r, 2025) \leq r < d$, contradicting the minimality of $d$. Thus $S$ is precisely the multiples of $d$.

It can be verified that for any $d \mid 2025$, the set of multiples of $d$ is balanced. Indeed, as $d$ is odd, for any $ad$, $bd \in S$, their average $\frac{(a + b)d}{2}$ is a multiple of $d$ and hence also in $S$. Thus, any shift of such a set is also balanced; as seen above, these classify all balanced sets. For each $d \mid 2025$, there are $d$ choices for $S$, so the answer is
$$\sum_{d \mid 2025} d = \frac{3^5 - 1}{2} \cdot \frac{5^3 - 1}{4} = \left\lceil \frac{3751}{2} \right\rceil.$$