1. Definitions and Setup:
- An integer $n$ is bad if $f(n) \neq n f(1)$.
- A pair of integers $(x, y)$ is bad if $f(x) + f(y) \neq f(x + y)$.
- A pair of integers $(x, y)$ is awful if at least one of $x$ or $y$ is bad and $x + y$ is also bad.

2. Interval Consideration:
- Fix an arbitrary $n > 0$ and consider the interval $I = [-n, n]$.
- Slightly tweak the definition of awful to require $x + y \in I$.

3. Counting Awful Pairs:
- For every bad $k \in I$, any $(x, y) \in I^2$ with $x + y = k$ must be either bad or awful.
- Since for any $k$ there are at least $n$ such pairs (actually $n + 1$, but we use $n$ for simplicity), if we have $k$ bad integers in $I$, then there are at least $nk - 0.001n^2$ awful pairs.

4. Bounding the Number of Bad Integers:
- It follows that there's some bad integer $B$ that's in at least $n - \frac{0.001n^2}{k}$ awful pairs.
- Since every awful pair generates a different bad integer as its sum, we have:
$$n - \frac{0.001n^2}{k} \leq k$$
Simplifying this inequality:
$$k^2 - nk + 0.001n^2 \geq 0$$
Solving the quadratic inequality:
$$k \in \left[0, \frac{n(50 - \sqrt{2490})}{100}\right] \cup \left[\frac{n(50 + \sqrt{2490})}{100}, \infty\right)$$

5. **Conclusion for Small $n$:**
- For small $n$ (say, $n \leq 30$), we actually need $f(x) + f(y) = f(x + y)$ for all $x, y \in [-n, n]$, which implies $f$ is linear over $I$.
- Thus, $k$ falls in the first interval (since it's zero), and if at some point we transition between $k$ being in the first interval to the second interval for $n \geq 30$, we have to increase $k$ by more than 2 by discrete continuity, which is impossible.

6. Final Conclusion:
- Therefore, $k$ stays in the first interval, so there are at most $n$ bad numbers in $[-n, n]$ (actually way less), as desired.

$\blacksquare$