1. Count all 4-digit numbers

The 4-digit numbers run from $1000$ to $9999$.

So, total number of 4-digit numbers is $9999-1000+1=9000.$

2. Count 4-digit multiples of $3$

We need multiples of $3$ between $1000$ and $9999$.

• First multiple of $3$ greater than or equal to $1000$ is $1002$.
• Last multiple of $3$ less than or equal to $9999$ is $9999$.

This is an AP: $1002, 1005, 1008, \dots, 9999$ with common difference $3$.

Number of terms is $\frac{9999-1002}{3}+1=\frac{8997}{3}+1=2999+1=3000.$

So, number of 4-digit multiples of $3$ is $3000.$

3. Count 4-digit multiples of $7$

We need multiples of $7$ between $1000$ and $9999$.

• First multiple of $7$ greater than or equal to $1000$ is $1001=7\times 143$.
• Last multiple of $7$ less than or equal to $9999$ is $9996=7\times 1428$.

Thus the multiples are $1001, 1008, 1015, \dots, 9996$

Number of such multiples is $1428-143+1=1286.$

So, number of 4-digit multiples of $7$ is $1286.$

4. Count 4-digit multiples of both $3$ and $7$

A number divisible by both $3$ and $7$ must be divisible by $\operatorname{lcm}(3,7)=21.$

We need 4-digit multiples of $21$.

• First multiple of $21$ greater than or equal to $1000$ is $1008=21\times 48$.
• Last multiple of $21$ less than or equal to $9999$ is $9996=21\times 476$.

Hence the number of 4-digit multiples of $21$ is $476-48+1=429.$

5. Apply inclusion-exclusion principle

Number of 4-digit numbers that are multiples of $3$ or $7$ is $3000+1286-429=3857.$

Therefore, the numbers that are neither multiples of $3$ nor multiples of $7$ are $9000-3857=5143.$

6. Final Answer

$\boxed{5143}$

The derived answer matches the stored correct answer.