Solution:

Let us construct infinitely many digitally divisible integers.

Consider the integer $111\ldots 1$ consisting of $k$ digits, all of which are $1$ (for any $k \geq 1$). None of its digits is zero, so condition (a) is satisfied.

The sum of its digits is $k$ (since there are $k$ digits, each $1$). The number itself is $N = \underbrace{111\ldots 1}_{k\text{ times}} = \frac{10^k - 1}{9}$.

We claim that $N$ is divisible by $k$ for infinitely many $k$.

But even if $k$ does not always divide $N$, we can construct other numbers. For example, consider the number $N = 1\underbrace{00\ldots 0}_{k-1\text{ times}}1\underbrace{00\ldots 0}_{k-1\text{ times}}1\ldots 1$ (with $n$ blocks of $1$ separated by $k-1$ zeros), i.e., $N = 10^{(n-1)k} + 10^{(n-2)k} + \cdots + 10^0$. The sum of its digits is $n$ (since there are $n$ ones and the rest are zeros). But this number contains zeros, so it does not satisfy condition (a).

Instead, consider the following construction:

Let $d$ be any positive integer not divisible by $10$ (so that its digits are all nonzero). Let $N = d \times 10^m + d \times 10^{2m} + \cdots + d \times 10^{km}$, where $m$ is chosen so that $d$ has $m$ digits and none of the digits of $d$ is zero. Then $N$ will have only the digits of $d$ repeated, so none of its digits is zero.

Alternatively, consider the following:

Let $n$ be any positive integer. Consider the number $N = 1\underbrace{11\ldots 1}_{n-1\text{ times}}$ (i.e., $n$ digits, all $1$). The sum of its digits is $n$. $N = \frac{10^n - 1}{9}$. For $n = 9$, $N = 111111111$, sum of digits is $9$, and $111111111$ is divisible by $9$.

In general, for $n = 9k$, $N = \underbrace{111\ldots 1}_{9k\text{ times}}$ is divisible by $9$, and the sum of its digits is $9k$, so $N$ is divisible by the sum of its digits. Thus, for every $k \geq 1$, $N = \underbrace{111\ldots 1}_{9k\text{ times}}$ is digitally divisible.

Therefore, there are infinitely many digitally divisible integers.