Solution:
$3232 = 16 \times 202$ and $10000 = 16 \times 625$. So any number with $3232$ as its last $4$ digits is divisible by $16$. So consider $N = 22223232$. Its sum of digits is $18$, so it is divisible by $9$. Hence it is divisible by $9 \cdot 16 = 144$. But any four digits have at most four $2$'s and at most two $3$'s, so the product of any four digits divides $144$ and hence $N$. But now we can extend $N$ by inserting an additional $9m$ $2$'s at the front. Its digit sum is increased by $18m$, so it remains divisible by $144$ and it is still divisible by the product of any four digits.

Solution:
The number $111111111$ with nine $1$'s is divisible by $9$. Hence the number with twenty-seven $1$'s, which equals $111111111 \times 1000000001000000001$, is divisible by $27$. So $N$, the number with twenty-seven $3$'s, is divisible by $3^{4}$. Now the number with $27n$ $3$'s is divisible by $N$ and hence by $3^{4}$.