Let $\displaystyle x_{1},x_{2},x_{3},x_{4}$ be the four digits of a four digit number $\displaystyle N=x_{1}x_{2}x_{3}x_{4}.$
Find the maximum value of $\displaystyle \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$
Wouldn't it be 1000?
$\displaystyle \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} \leq N$
So, if $\displaystyle x_{1}+x_{2}+x_{3}+x_{4} = 1$, then we get $\displaystyle N=N$, which is the most that $\displaystyle \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$ can be.
Since the number must have four digits, we can't have $\displaystyle x_{1} = 0$, so it must be at least 1, and since $\displaystyle x_{1}+x_{2}+x_{3}+x_{4} = 1$, we have that $\displaystyle x_{2}=x_{3}=x_{4} = 0$.
So, we get $\displaystyle N=1000$ and $\displaystyle \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000$.
Actually, N=2000,3000,4000,...,9000 works also, but in all those cases we have that $\displaystyle \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000$
$\displaystyle N\ =\ 1000x_1+100x_2+10x_3+x_4$
$\displaystyle \le\ 1000x_1+1000x_2+1000x_3+1000x_4$
$\displaystyle \therefore\ \frac N{x_1+x_2+x_3+x_4}\ \le\ 1000$
Equality is attained when $\displaystyle x_2=x_3=x_4=0.$ Hence the maximum value of $\displaystyle N$ is 1000.