SAUDI ARABIAN IMO Booklet, 2023

The smallest such number is $n = 9$.
Suppose that $n$ is a number allowing for coloring with the desired properties. Then $n \ge 2$ since each color needs to be used at least once. Note that $n$ cannot be blue since it is inexpressible as a difference of two numbers between $1$ and $n$, therefore $n$ is red. On the other hand, $1$ and $2$ cannot be red, since none of them is a sum of two distinct positive integers, which means $1$ and $2$ are blue (and in particular $n \ge 3$). Now $n-1$ cannot be a difference of two red numbers (since $1$ is blue), so $n-1$ is red. Similarly $n-2 = (n-1)-1$ has to be red since both $1$ and $2$ are blue. It follows that $n \ge 5$.

Since $n$ is red, it needs to be a sum of two distinct blue numbers. But a sum of two blue numbers cannot exceed
$$(n-3) + (n-4) = 2n-7.$$
Therefore, $2n - 7 \ge n \implies n \ge 7$.
If $n = 7$, then the only remaining possibility to express $7$ as a sum of two blue numbers is $3+4$, but then $4$ is not a difference of two red numbers, contradiction. Similarly with $n = 8$.
We have proved that $n \ge 9$. It remains to show that there exists a valid coloring for $n = 9$. We can color $1$, $2$, $4$, $5$, $6$ blue and $3$, $7$, $8$, $9$ red. Correctness of this coloring follows from the simple arithmetic:
$$1 = 9 - 8,\ 2 = 9 - 7,\ 4 = 7 - 3,\ 5 = 8 - 3,\ 6 = 9 - 3,$$
$$3 = 1 + 2,\ 7 = 1 + 6,\ 8 = 2 + 6,\ 9 = 4 + 5.$$