Let $x$ and $y$ be two positive integers such that $x+y=2310$. Suppose $xy$ is divisible by $2310$, then let $xy=2310n$ where $n$ is a positive integer. Then,

$$x+\Big(2310\cdot\frac{n}{x}\Big)=2310 \implies x^2-2310x+2310n=0$$

is solvable in integers. This means its discriminant $$\Delta=2310^2-4\times(2310n)=2^2\times 3\times 5\times 7\times 11\times (1155-2n)$$ must be a perfect square. If so, $$1155-2n=3\times 5\times 7\times 11\times k^2 \ge 1155$$ which cannot be true.