The function $f: \mathbb{N}\to\mathbb{N}_{0}$ satisfies for all $m,n\in\mathbb{N}$: \[f(m+n)-f(m)-f(n)=0\text{ or }1, \; f(2)=0, \; f(3)>0, \; \text{ and }f(9999)=3333.\] Determine $f(1982)$.
1. **Given Conditions and Initial Analysis:**
   - The function \( f: \mathbb{N} \to \mathbb{N}_0 \) satisfies:
     \[
     f(m+n) - f(m) - f(n) = 0 \text{ or } 1
     \]
     for all \( m, n \in \mathbb{N} \).
   - Additional conditions:
     \[
     f(2) = 0, \quad f(3) > 0, \quad \text{and} \quad f(9999) = 3333.
     \]

2. **Determining \( f(1) \):**
   - Assume \( f(1) = k \).
   - Using \( f(2) = 0 \):
     \[
     f(2) = f(1+1) = f(1) + f(1) + c \quad \text{where} \quad c \in \{0, 1\}
     \]
     \[
     0 = k + k + c \implies 0 = 2k + c
     \]
     Since \( c \) can be 0 or 1, and \( k \) must be a non-negative integer:
     \[
     k = 0 \quad \text{and} \quad c = 0
     \]
     Therefore, \( f(1) = 0 \).

3. **Determining \( f(3) \):**
   - Using \( f(3) > 0 \):
     \[
     f(3) = f(2+1) = f(2) + f(1) + c = 0 + 0 + c \implies f(3) = c
     \]
     Since \( f(3) > 0 \), \( c \) must be 1:
     \[
     f(3) = 1
     \]

4. **Inductive Step for \( f(3n) \):**
   - Assume \( f(3n) = n \) for some \( n \).
   - For \( f(3(n+1)) \):
     \[
     f(3(n+1)) = f(3n + 3) = f(3n) + f(3) + c
     \]
     \[
     f(3(n+1)) = n + 1 + c
     \]
     Since \( c \) can be 0 or 1, and we need \( f(3(n+1)) = n+1 \):
     \[
     c = 0
     \]
     Therefore, by induction:
     \[
     f(3n) = n \quad \text{for all} \quad n \leq 3333
     \]

5. **Determining \( f(3n+1) \) and \( f(3n+2) \):**
   - For \( f(3n+1) \):
     \[
     f(3n+1) = f(3n) + f(1) + c = n + 0 + c
     \]
     Since \( f(3n+1) \) must be an integer and \( c \) can be 0 or 1:
     \[
     f(3n+1) = n
     \]
   - For \( f(3n+2) \):
     \[
     f(3n+2) = f(3n) + f(2) + c = n + 0 + c
     \]
     Since \( f(3n+2) \) must be an integer and \( c \) can be 0 or 1:
     \[
     f(3n+2) = n
     \]

6. **Calculating \( f(1982) \):**
   - Since \( 1982 = 3 \times 660 + 2 \):
     \[
     f(1982) = f(3 \times 660 + 2) = f(3 \times 660) + f(2) + c = 660 + 0 + c
     \]
     Since \( c \) can be 0 or 1, and \( f(1982) \) must be an integer:
     \[
     f(1982) = 660
     \]

The final answer is \( \boxed{660} \).