Answer. $n=208,304$ Solution. For the given expression to be a divisor of $n$, it should be an integer. Therefore for a $t\in\mathbb{Z}$, $$\sqrt{4n+9}=2t+1\Rightarrow n=\frac{(2t+1)^2-9}{4}=t^2+t-2=(t-1)(t+2)$$must hold. Moreover, $$\frac{3+\sqrt{4n+9}}{2}=t+2$$is true. When checked, we see that neither $t$ nor $t+1$ can divide $n$. Due to this, we obtain the fact that $t-1$ is the fifth smallest positive divisor of $n$. So $n$ must have exactly $10$ divisors. Thus, there are two possibilities: $(i)$ $n=p^9$ when $p$ is a prime. From this we get $(t+2)-(t-1)=3\Rightarrow p^4\mid 3$, which is a contradiction. $(ii)$ $n=pq^4$ when $p$ and $q$ are distinct prime numbers. Observe that $(t-1,t+2)=d\in\{1,3\}$ is true. We seperate this to two cases. $d=1\Rightarrow \{p,q^4\}=\{t+2,t-1\}$ must hold. From this we get $\mid p-q^4\mid=3$. Due to parity, one of $p$ and $q$ must equal to $2$. Therefore the only solutions to this equation are $(13,2),(19,2)$. These give out the aforementioned solutions. $d=3\Rightarrow 9\mid n\Rightarrow q=3\Rightarrow n=81p.$ When checked manually it can be seen that $p=2,3,5,7,11$ do not hold. Due to this $p$ must be greater than or equal to $13$. We seperate this into 3 cases. $13\leq p<27\Rightarrow p=10$, contradiction. $27\leq p<81\Rightarrow 2p=3$, contradiction. $p>81\Rightarrow p=84$, contradiction. Ergo, there are no solutions besides $208$ and $304$ since we have exhausted all the cases. $\blacksquare$