HMMT February

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.