HMMT November, 2024

Solution:
Denote the digit sum of a positive integer $m$ as $s(m)$.
Let the pile have $n$ gypsum crystals, so that $n$ can be written as $10^{i_{1}} + 10^{i_{2}} + \cdots + 10^{i_{s(n)}}$.

First, $s(n)$ cannot be $1$ (i.e. $n$ cannot be a power of $10$), since otherwise we could split the gypsum pile into two equal parts each with digit sum $5$. We also can't have $1 < s(n) \leq 12$, as otherwise $n = 10^{i_{1}} + 10^{i_{2}} + \cdots + 10^{i_{s(n)}}$ can be split into two groups of at most six terms each (which have digit sum at most $6$). Hence $s(n)$ is at least $13$, forcing $n$ to be at least $49$.

Now we show that $n = 49$ indeed works. Note that for any splitting into two piles of size $a, b$ with $a + b = 49$, we do not carry any digits when adding $a$ and $b$. Hence we must have $s(a) + s(b) = s(n) = 13$. This forces at least one of $s(a), s(b)$ to be at least $7$, so $n = 49$ indeed works.