Let the 4-digit number be $n$.

We need $\gcd(n,18)=3.$

Factorize $18$: $18=2\cdot 3^2.$

For $\gcd(n,18)$ to be exactly $3$, the number $n$ must satisfy:

• it is divisible by $3$,
• it is not divisible by $2$,
• it is not divisible by $9$.

Why?

• Since the gcd contains a factor $3$, $n$ must be divisible by $3$.
• If $n$ were even, then gcd would also contain $2$, making it at least $6$.
• If $n$ were divisible by $9$, then gcd would contain $9$, making it $9$ (or $18$ if also even), not $3$.

So we count 4-digit numbers from $1000$ to $9999$ that are: $\text{divisible by }3, \quad \text{not divisible by }2, \quad \text{not divisible by }9.$

First count 4-digit multiples of $3$.

Smallest 4-digit multiple of $3$ is $1002$.

Largest 4-digit multiple of $3$ is $9999$.

Number of terms in this arithmetic progression: $\frac{9999-1002}{3}+1=2999+1=3000.$

Among these, count those also divisible by $2$, i.e. divisible by $6$.

Smallest 4-digit multiple of $6$ is $1002$.

Largest 4-digit multiple of $6$ is $9996$.

Count: $\frac{9996-1002}{6}+1=1499+1=1500.$

Next count those divisible by $9$.

Smallest 4-digit multiple of $9$ is $1008$.

Largest 4-digit multiple of $9$ is $9999$.

Count: $\frac{9999-1008}{9}+1=999+1=1000.$

But numbers divisible by both $6$ and $9$ are divisible by $\operatorname{lcm}(6,9)=18.$

Smallest 4-digit multiple of $18$ is $1008$.

Largest 4-digit multiple of $18$ is $9990$.

Count: $\frac{9990-1008}{18}+1=499+1=500.$

By inclusion-exclusion, required count is $3000-1500-1000+500=1000.$

Therefore, the total number of 4-digit numbers with $\gcd(n,18)=3$ is $\boxed{1000}.$