HMMT February, 2019

Solution:

We claim that all the weird numbers are all the prime numbers and $4$.

Since no numbers between $1$ and $p-2$ divide prime $p$, $(p-2)!$ will not be divisible by $p$. We also have $2! = 2$ not being a multiple of $4$.

Now we show that all other numbers are not weird. If $n = pq$ where $p \neq q$ and $p, q \geq 2$, then since $p$ and $q$ both appear in $1, 2, \ldots, n-2$ and are distinct, we have $pq \mid (n-2)!$. This leaves the only case of $n = p^2$ for prime $p \geq 3$. In this case, we can note that $p$ and $2p$ are both less than $p^2 - 2$, so $2p^2 \mid (n-2)!$ and we are similarly done.

Since there are $25$ prime numbers not exceeding $100$, there are $25 + 1 = 26$ weird numbers.