Find two numbers whose sum is 20 and LCM is 24

With 4th grade knowledge in mind, I would say this question is about finding all the factors of $24$, because the two numbers must both be factors of $24$ or they would have a different least common multiple.

So listing the factors of $24$ we get $1,2,3,4,6,8,12,24$. Then it's easy to see which two add to $20$, although you could certainly use the exercise, if so inclined, to develop a few more rules for harder cases, and in another case you might need to be a little careful that the least common multiple is indeed still $24$.

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Another approach: $\frac{20}{24} = \frac{5}{6}$ - can we write $\frac{5}{6}$ as the sum of two unit fractions? The answer is yes, $\frac{1}{2} + \frac{1}{3}$, and the numbers we seek are $\frac{24}{2}=12$ and $\frac{24}{3}=8$

A general algorithm

If $a+b = n$ and $\mathrm{lcm}(a,b) = c$ we let $\gcd(a,b) = d$ to get $$ab = cd\\ n = a+b$$ Solving $b = n-a$ gives us $$a(n-a) = cd \\ \Leftrightarrow a^2-na + cd = 0 \\ \Leftrightarrow a = \frac n2 \pm \sqrt{\frac{n^2}4 - cd}$$ So for even $n$ we must find $d$ such that $\frac{n^2}4 - cd$ is a perfect square (since $c>0$ this will amount to a finite number of possibilities). The $\pm$ is irrelevant because $b$ will take the alternate value.
If $n$ is odd, $n^2 - 4cd$ must be a perfect square instead and we obtain an analogous formula: $$a = \frac12 (n \pm \sqrt{n^2 - 4cd})$$

In both cases, $d$ can range between $1$ and $\left\lfloor \frac{n^2}{4c} \right\rfloor$

Your case thus allows $1\le d\le \lfloor\frac{400}{96}\rfloor = 4$ and $100-24d$ must be a perfect square. $d=4$ yields $4 = 2^2$ so $$a = 10 + \sqrt{4} = 12; \quad b = 10 - \sqrt4 = 8$$

With an LCM of 24 one of the numbers must be 8, as $2^3$ divides 24. It can't be 16 as 16 does not divide 24.