Since the product of the digits of $N$ is 1728, we find the prime factorization of 1728 to help us determine what the digits are: $1728=9 \times 192=3^{2} \times 3 \times 64=3^{3} \times 2^{6}$. We must try to find a combination of the smallest number of possible digits whose product is 1728. Note that we cannot have 3 digits with a product of 1728 since the maximum possible product of 3 digits is $9 \times 9 \times 9=729$. Let us suppose that we can have 4 digits with a product of 1728. In order for $N$ to be as small as possible, its leading digit (that is, its thousands digit) must be as small as possible. From above, this digit cannot be 1. This digit also cannot be 2, since otherwise the product of the remaining 3 digits would be 864 which is larger than the product of 3 digits can be. Can the thousands digit be 3? If so, the remaining 3 digits have a product of 576. Can 3 digits have a product of 576? If one of these 3 digits were 7 or less, then the product of the 3 digits would be at most $7 \times 9 \times 9=567$, which is too small. Therefore, if we have 3 digits with a product of 576, then each digit is 8 or 9. Since the product is even, then at least one of the digits would have to be 8, leaving the remaining two digits to have a product of $576 \div 8=72$. These two digits would then have to be 8 and 9. Thus, we can have 3 digits with a product of 576, and so we can have 4 digits with a product of 1728 with smallest digit 3. Therefore, the digits of $N$ must be $3,8,8,9$. The smallest possible number formed by these digits is when the digits are placed in increasing order, and so $N=3889$. The sum of the digits of $N$ is $3+8+8+9=28$.