Team Selection Test for JBMO

The answer is $1$, $5$ and $8$.

Checking by hand for $n=1,2,\dots ,6$, we see that $1$ and $5$ work. For $n\ge 7$, $2n+7$ should be a prime number. Because, otherwise there exists a prime divisor of $2n+7$ which is less than or equal to $n$ since $2n+7$ is odd, but it divides $n!$.

Now let $2n+7=p\ge 21$ where $p$ is a prime number. Then the condition is equivalent to $\left(\frac{p-7}{2}\right)!\equiv 1(\mathrm{mod}p)$. Wilson's theorem gives that $(p-1)!\equiv -1(\mathrm{mod}p)$. On the other hand $$(p-1)!\equiv (-1{)}^{\frac{p-7}{2}}\cdot \left(\frac{p-7}{2}\right){!}^2\cdot \frac{p-5}{2}\cdot \frac{p-3}{2}\cdot \frac{p-1}{2}\cdot \frac{p+1}{2}\cdot \frac{p+3}{2}\cdot \frac{p+5}{2}\equiv (-1{)}^{\frac{p-1}{2}}\frac{225}{64}(\mathrm{mod}p)$$ $$\text{Thus we obtain }225\equiv (-1{)}^{\frac{p+1}{2}}64(\mathrm{mod}p).$$ If $p\equiv 1(\mathrm{mod}4)$, then $p\mid 225+64=17^2$ but $p\ge 21$, no solution exists. If $p\equiv 3(\mathrm{mod}4)$, then $p\mid 225-64=7\cdot 23$. As $p\ge 21$ we have $p=23$, that is $n=8$ and it satisfies the condition.