The boys of a school can be arranged in $15$, $18$ or $24$ equal rows and also into a solid square. Find the least number of boys in the school.

First, let's find the LCM of $15$, $18$, and $24$:
* $15=3\times 5$
* $18=2\times 3^2$
* $24=2^3\times 3$

$$\text{LCM}=2^3\times 3^2\times 5=8\times 9\times 5=360$$

The prime factorization of the LCM is $2^3\times 3^2\times 5^1=2^2\times 2^1\times 3^2\times 5^1$.
For the arrangement to form a solid square, the number of boys must be a perfect square. Thus, we must multiply the LCM by the missing prime factors to complete the pairs (i.e., multiply by $2$ and $5$):

$$\text{Least number of boys}=360\times 2\times 5=3600$$