Problem 1596

We are given that the function $f(n)$ is defined on positive integers and it takes non-negative integer values. It satisfies:

$$f(2) = 0,$$
$$f(3) > 0,$$
$$f(9999) = 3333,$$

and for all $m, n$:

$$f(m+n) - f(m) - f(n) = 0 \text{ or } 1.$$

We need to determine $f(1982)$.

### Analysis of the Function $f(n)$

Given the functional equation:

$$f(m+n) = f(m) + f(n) \text{ or } f(m) + f(n) + 1,$$

we observe that $f(n)$ behaves much like an additive function with an additional constraint. Furthermore, the values provided imply a linear-like growth with periodic modifications due to the $+1$ term in the equation.

### Establishing a Hypothesis

1. Hypothesis of Linear Growth: Given that $f(9999) = 3333$, a reasonable first hypothesis for $f(n)$ is that it is approximately proportional to $n$, suggesting $f(n) \approx \frac{n}{3}$.

2. Discrete Steps with Deviations: The functional condition allows for deviations of $+1$ from strict linearity, indicating some periodic rate of adjustment.

### Verifying Consistency of $f(n)$

Using the assumption $f(n) = \left\lfloor \frac{n}{3} \right\rfloor$, let's verify with the given information:

- $f(2) = 0$: The formula $\left\lfloor \frac{2}{3} \right\rfloor = 0$ agrees.
- $f(3) > 0$: Indeed, $\left\lfloor \frac{3}{3} \right\rfloor = 1$ agrees.
- $f(9999) = 3333$: Indeed, $\left\lfloor \frac{9999}{3} \right\rfloor = 3333$ agrees.

### Calculating $f(1982)$

To find $f(1982)$:

$$f(1982) = \left\lfloor \frac{1982}{3} \right\rfloor$$

Carrying out the division:

$$\frac{1982}{3} = 660.666\ldots$$

Taking the floor function:

$$\left\lfloor \frac{1982}{3} \right\rfloor = 660$$

Thus, the value of $f(1982)$ is:

$$\boxed{660}$$