Problem 1784

1. Assume the existence of arbitrarily large red intervals:
Suppose there exist arbitrarily large red intervals. Pick any two red integers $a$ and $b$ and consider a red interval $[M, M+ab]$. We should have:
$$f(M) + af(b) = f(M+b) + (a-1)f(b) = \cdots = f(M+ab) = f(M+ab-a) + f(a) = \cdots = f(M) + bf(a)$$
Hence, $f(x) = cx$ for some $c > 0$ on the red integers. The symmetric result holds for blue intervals. Thus, if there exist arbitrarily large monochrome intervals of both colors, we are immediately done.

2. Assume only one color has arbitrarily large intervals:
If there only exist arbitrarily large monochromatic intervals of one color—WLOG red—then there exists some $t > 0$ such that blue intervals contain at most $t$ integers. Now, $f(x) \leq c(x+t)$ in general if $f(x) = cx$ over red $x$, which also finishes.

3. Assume bounded monochromatic intervals:
Suppose there exists some integer $T > 0$ such that monochromatic red or blue intervals contain at most $T$ integers.

4. Lemma:
For any $2T+1$ consecutive positive integers, there exist $r_1, r_2$ red and $b_1, b_2$ blue among them such that $r_1 + r_2 = b_1 + b_2$.

Proof:
If we can find two non-adjacent numbers of the same color (WLOG red) $a, b$ such that every $a < x < b$ is blue, we are done, since $a + b = (a+1) + (b-1)$ and $a+1$ and $b-1$ (not necessarily distinct) are blue by definition. If we split our $2T+1$ consecutive integers into maximal monochromatic "blocks", we need at least 3 blocks since each block is at most $T$ integers long. Hence, the rightmost number in the first block and the leftmost number in the third block will work. $\blacksquare$

5. **Define $g(n)$:**
For nonnegative integers $n$, let $g(n)$ denote the maximum of $\frac{f(t)}{t}$ for $1 \leq t \leq \lfloor 1.5^n \cdot 1000T \rfloor$. Clearly, it suffices to show that $g$ is bounded above. This will follow from the following key claim after some algebra.

6. Claim:
We have:
$$g(n+1) \leq g(0) \prod_{i=0}^n \left(1 + \frac{4T+1}{1000T} 1.5^{-i}\right)$$

Proof:
Pick some $\lfloor 1.5^n \cdot 1000T \rfloor < t \leq \lfloor 1.5^{n+1} \cdot 1000T \rfloor$. Let $m = \lceil t/2 \rceil$; by the lemma, there exist red $r_1, r_2$ and blue $b_1, b_2$ among $\{m, \ldots, m+2T\}$ such that $r_1 + r_2 = b_1 + b_2 := k$, so:
$$t \leq 2m \leq k \leq 2m + 4T \leq t + 4T + 1$$
Now, regardless of the color of $k$, we have:
$$\frac{f(k)}{k} \leq \max\left(\frac{f(r_1)}{r_1}, \frac{f(r_2)}{r_2}, \frac{f(b_1)}{b_1}, \frac{f(b_2)}{b_2}\right) \leq g(n)$$
since we have $m + 2T \leq \lfloor 1.5^n \cdot 1000T \rfloor$. Thus:
$$f(t) \leq f(k) \leq kg(n) \implies \frac{f(t)}{t} \leq \frac{k}{t} g(n) \leq \left(1 + \frac{4T+1}{1.5^n \cdot 1000T}\right) g(n)$$
So $g(n+1)$ is at most the RHS since $t$ is arbitrarily chosen. This finishes by induction. $\blacksquare$

7. **Bound $g(n)$:**
Since $g(0)$ is finite, it suffices to show that:
$$\prod_{i=0}^\infty \left(1 + \frac{4T+1}{1000T} 1.5^{-i}\right)$$
converges. Since every term is at least 1, by taking logs it suffices to show that:
$$\sum_{i=0}^\infty \ln \left(1 + \frac{4T+1}{1000T} 1.5^{-i}\right)$$
converges. Using the well-known inequality $\ln(1+x) \leq x$, it suffices to show that:
$$\sum_{i=0}^\infty \frac{4T+1}{1000T} 1.5^{-i}$$
converges, which is obviously true. $\blacksquare$

The final answer is $\boxed{ f(x) \leq ax }$ for some positive number $a$.