There Can Be Infinitely Many

Consider $\max\{\text{odd} ~ a:a^2< n\}$ . For $a\ge 7$ , $n$ must be divisible by all of $a, a-2,a-4$ . Notice that $\gcd(a,a-2)=\gcd(a-2,a-4)=\gcd(a,a-4)=1$ so $a(a-2)(a-4)\mid n$ . Now clearly $n\le (a+2)^2$ . So we have $a(a-2)(a-4)\le n\le (a+2)^2\implies 4(a-1)+a^2(a-7)\le 0$ which is clearly not true for $a\ge 7$ . Hence $a\in\{1,3,5\}$ .

If $a=1$ then $1\le n\le 9$ so $n\in\{1,2,\cdots,9\}$ .

If $a=3$ then $9\le n \le 25$ and $3\mid n$ so $n\in\{9,12,\cdots,24\}$ .

If $a=5$ then $25\le n\le 49$ and $15\mid n$ so $n\in\{30,45\}$ .

The solution set is $\{1,2,\cdots,8,9\}\cup \{9,12,\cdots,24\}\cup\{30,45\}$ .

The sum is $\boxed{210}.$

From the problem description, the set of factors of a sacred number n must be a superset of a set of the form S= {1, 3, 5, ... 2k-1}. Suppose it contains {1,3,5,7}. Then n >= 105. So S contains 9 too. But S contains {1,3,5,7,9} implies n>=315 so S contains {1,3,5,7,9,11,13,15,17}, and so on. In other words, S cannot contain 7. In other words, n < 49. Now brute force your way through the 48 possibilities to find that [1 .. 9, 12, 15, 18, 21, 24, 30, 45] are the only possibilities.

If $k^2\le n<(k+2)^2$ where $k$ is an odd integer and $n$ is sacred , then $\mbox{LCD}(1,3,...,k)|n$ , which implies that: $(k+2)^2>n>\mbox{LCD}(1,3,...,k)$

Observe that $\mbox{LCD}(1,3,...,k)$ is larger than the product of the primes in that list, making the inequality into: $(k+2)^2>n>\mbox{LCD}(1,3,...,k)>\mbox{(product of primes)}$

Observe that when $k=11$ , the product of odd primes equal to or below 11 is $1155$ , which is larger than $13^2$ , which sets the upper boundary of sacred number to 121.

When $1^2\le n<3^2$ , the LCD is 1 and the sacred numbers are 1 to 8, with a sum of 36.

When $3^2\le n<5^2$ , the LCD is 3 and the sacred numbers are 9,12,15,18,21, and 24, with a sum of 99.

When $5^2\le n<7^2$ , the LCD is 15 and the sacred numbers are 30 and 45, with a sum of 45.

There is no sacred numbers beyond $7^2$ .

Therefore the required value is $36+99+45=\fbox{210}$ .

(Is it a coincidence that the points awarded for this question is 210?)