HMMT February, 2022

Solution:
Note that from $n = m^2$ to $n = (m+1)^2$, $\|n\|$ increases from $0$ to a peak of $m$ (which is repeated twice), and then goes back down to $0$. Therefore
$$\sum_{n=1}^{m^2} \|n\| = \sum_{k=1}^{m-1} 2(1+2+\cdots+k) = \sum_{k=1}^{m-1} 2\binom{k+1}{2} = 2\binom{m+1}{3} = \frac{m}{3}\left(m^2-1\right)$$
In particular, if $n = m^2 - 1$,
$$\frac{\|1\|+\|2\|+\cdots+\|n\|}{n} = \frac{m}{3},$$
so $n = 300^2 - 1$ satisfies the condition. However, this does not prove that there are not smaller solutions for $n$.
Let $N = 300^2 - 1$ and suppose that $N - k$ satisfies the condition. Then, we know that
$$\frac{\|N\| + \|N-1\| + \cdots + \|N-(k-1)\|}{k} = 100.$$
Since $\|N-k\| = k+1$ for $k \leq 298$, one can show that $k = 199$ works. By looking at further terms, one can convince oneself that no larger value of $k$ works. Thus, the answer is $300^2 - 1 - 199 = 90000 - 200 = 89800$.