Divisors!

Let, $n=2^{k} \prod p_{i}^{q_{i}}$ where $p_{i }$ is any particular prime number excluding 2. Now, $d(n)=(k+1) \prod (q_{i}+1)$ The numbers we are trying to find must satisfy $d(n)=\frac{n}{2}$ or, $(k+1) \prod (q_{i }+1)=2^{k-1} \prod p_{i }^{q_{i }}$ or, $\frac{k+1}{2^{k-1}}=\frac{\prod p_{i }^{q_{i }}}{\prod (q_{i }+1)}$ ...(1)

Now, $\frac{\prod p_{i }^{q_{i }}}{\prod (q_{i }+1)} \geq 1$ where equality occurs only when all $q_{i }=0$ . Hence, the following must be true. $\frac{k+1}{2^{k-1}} \geq 1$ This happens for $k \leq 3$ . As we can see $k=3$ makes both sides of equation (1) equal to 1. And, this corresponds to the number $2^3=8$ . $k=2$ makes the LHS side equal to $\frac{3}{2}$ . So, we need $\prod p_{i }^{q_{i }}=3^1$ . This time the number is $2^2 3^1=12$ . $k=1$ demands the RHS should be equal to 2. But, we cannot grind out a 2 from $\prod p_{i }^{q_{i }}$ as it's the product of primes excluding 2. Therefore, $k=1$ doesn't give us a new number. Finally, the answer is $8+12=\boxed {20}$ .

HINT: The number of divisors of a natural number $n \leq \lfloor\sqrt{n}\rfloor \times 2$ .