The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from $0$ to this maximum limit of $80\:\mathrm{N}$ that you found.
In math-terms you have looked at static friction $f_s$ with this expression:
$$f_s\leq n \mu_s$$
There is only an equal sign here, if you are looking for maximum static friction. If you are just looking for static friction, you cannot use this. Always, when you have a force that you don't have a formula for, then use Newtons laws to find it:
Newtons 2nd law on the top block:
$$\sum F_x=m_{top}a\quad\Leftrightarrow\quad F-f_s=m_{top}a\quad\Leftrightarrow\quad f_s=F-m_{top}a$$
Newtons 2nd law on the bottom block:
$$\sum F_x=m_{bottom}a\quad\Leftrightarrow\quad f_s=m_{bottom}a \quad\Leftrightarrow\quad a=\frac{f_s}{m_{bottom}}$$
The accelerations $a$ are equal if this is static friction (if they are kept together). This is two equations with only two unknowns. If you put in numbers and solve them you get the right result.
You have in your question shown that the maximum static friction is $80\:\mathrm{N}$, so if our result here is larger, then we know that we do not have static but rather kinetic friction between them. Then you could easily have found the result without Newtons laws. Because in that case it must be kinetic friction $f_k$ between the boxes (as this is the only other possibility) since they slide over each other. And you do have a clear formula for kinetic friction:
$$f_k=n\mu_k$$
Note that this is another $\mu$ than before (typically $\mu_k$ is smaller than $\mu_s$)
