Answer: 890

Small Hint:

Reduce $n^3+100$ modulo $n+10$

Big Hint:

The condition makes $n+10$ a positive divisor of a fixed integer

Solution:

Modulo $n+10,$ we have $n\equiv-10.$ Thus $$\begin{gathered} n^3+100 \equiv(-10)^3+100\\ \equiv-900\pmod{n+10}. \end{gathered}$$ The condition is therefore equivalent to $900$ being divisible by $n+10.$ Since $n$ is positive, $n+10$ is a positive divisor of $900,$ and its largest possible value is $900.$ This gives $n=900-10=890.$