1. We need to count 4-digit numbers $\le 2800$ that are divisible by $3$ or by $11$.

2. Since 4-digit numbers start from $1000$, the required set is: $1000 \le n \le 2800$

We use inclusion-exclusion: $N(3 \text{ or } 11)=N(3)+N(11)-N(33)$ because numbers divisible by both $3$ and $11$ are divisible by $\mathrm{lcm}(3,11)=33$.

Step 1: Count numbers divisible by 3

We want multiples of $3$ in $[1000,2800]$.

• Smallest multiple of $3$ greater than or equal to $1000$: $\lceil 1000/3 \rceil = 334 \Rightarrow 334\cdot 3=1002$
• Largest multiple of $3$ less than or equal to $2800$: $\lfloor 2800/3 \rfloor = 933 \Rightarrow 933\cdot 3=2799$

So the multiples are: $1002,1005,\dots,2799$ Number of terms: $933-334+1=600$ Hence, $N(3)=600$

Step 2: Count numbers divisible by 11

We want multiples of $11$ in $[1000,2800]$.

• Smallest multiple of $11$ greater than or equal to $1000$: $\lceil 1000/11 \rceil = 91 \Rightarrow 91\cdot 11=1001$
• Largest multiple of $11$ less than or equal to $2800$: $\lfloor 2800/11 \rfloor = 254 \Rightarrow 254\cdot 11=2794$

Number of such multiples: $254-91+1=164$ Hence, $N(11)=164$

Step 3: Count numbers divisible by both 3 and 11

These are multiples of $33$ in $[1000,2800]$.

• Smallest multiple of $33$ greater than or equal to $1000$: $\lceil 1000/33 \rceil = 31 \Rightarrow 31\cdot 33=1023$
• Largest multiple of $33$ less than or equal to $2800$: $\lfloor 2800/33 \rfloor = 84 \Rightarrow 84\cdot 33=2772$

Number of such multiples: $84-31+1=54$ Hence, $N(33)=54$

Step 4: Apply inclusion-exclusion

$N(3 \text{ or } 11)=N(3)+N(11)-N(33)$ $=600+164-54$ $=710$

Final Answer

The number of 4-digit numbers less than or equal to $2800$ and divisible by $3$ or $11$ is: $\boxed{710}$