The smallest number which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is:

The required smallest number can be found using the formula:
$$\text{Number}=\text{LCM}(12,15,20,54)+\text{Remainder}$$

Let's find the prime factorization of each number:

$$12=2^2\times 3$$

$$15=3\times 5$$

$$20=2^2\times 5$$

$$54=2\times 3^3$$

Taking the highest powers of all prime factors involved to find the Least Common Multiple (LCM):

$$\text{LCM}=2^2\times 3^3\times 5=4\times 27\times 5=540$$

Adding the constant remainder of $4$:

$$\text{Required Number}=540+4=544$$