Let's do casework:

• $1^2 \le n < 3^2$, i.e. $a=1$, then $1|n \implies n = 1, 2, 3, 4, 5, 6, 7, 8$
• $3^2 \le n < 5^2$, i.e. $a=1, 3$, then $3|n \implies n = 9, 12, 15, 18, 21, 24$
• $5^2 \le n < 7^2$, i.e. $a=1, 3, 5$. Because $3$ and $5$ are co-prime, we must have $15|n \implies n = 30, 45$
• $7^2 \le n$, then there is no solution. This is because by # 2218, we have $$k \ge 4 \implies lcm(1, 3,\cdots, 2k- 3, 2k- 1) > (2k + 1)^2$$

In this case, because $1$, $3$, $\cdots$, $a=2k-1$ all divide $n$, thus their least common multiple must divide $n$ too. However, the above inequality shows that their least common multiple is greater than $(2k+1)^2 = (a+2)^2 > n$. Hence, it is impossible.