We need the highest power of $3$ dividing $66!$.

This is found using Legendre’s formula:

$v_3(66!)=\left\lfloor \frac{66}{3} \right\rfloor+\left\lfloor \frac{66}{3^2} \right\rfloor+\left\lfloor \frac{66}{3^3} \right\rfloor+\left\lfloor \frac{66}{3^4} \right\rfloor+\cdots$

Now compute each term:

1. $\left\lfloor \frac{66}{3} \right\rfloor=22$
2. $\left\lfloor \frac{66}{9} \right\rfloor=7$
3. $\left\lfloor \frac{66}{27} \right\rfloor=2$
4. $\left\lfloor \frac{66}{81} \right\rfloor=0$

So,

$v_3(66!)=22+7+2=31$

Hence, the largest natural number $n$ such that $3^n$ divides $66!$ is

$\boxed{31}$

Comparison with stored answer

Stored correct answer = $31$.

Our derived answer also is $31$, so they agree.