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 $24$.

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.

You want two factors of $24$ that add up to $20$. That means one factor must be less than or equal to $10$ (i.e. $20/2$), and the other must be between $10 $ and $20$ (inclusive) since it is equal to the first subtracted from $20$.

The only factor of $24$ between $10$ and $20$ inclusive is $12$. $20-12 = 8$. $8$ is a factor.

So the solution is $8$ and $12$.

Let $x$ and $y$ be the numbers. Note that $xy$ is the product of the LMC and the gcd, say, $d$. Then: $$xy=24d$$ $$x+y=20$$

I don't know if at 4th grade the pupils are supposed to be able of solve systems. If it is the case, you can write $$y=20-x$$ and then $$x(20-x)=24d$$

Solve $x$ to get $$x=\frac{20\pm\sqrt{400-96d}}2=10\pm2\sqrt{25-6d}$$

Therefore, $25-6d$ must be a perfect square. This only happens if $d=4$, which yields $x=8$ or $x=12$.

This is not a formula, but it makes the search much easier.