Solution:

The sum of the digits of $1996$ is $25$ and the sum of the digits of $2 \cdot 1996 = 3992$ is $23$. Because $1996 = 78 \cdot 25 + 46$, the number obtained by writing $78$ $1996$'s and two $3992$ in succession satisfies the condition of the problem.

As $3 \cdot 1996 = 5988$, the sum of the digits of $5988$ is $30$, and $1996 = 65 \cdot 30 + 46$, the number $39923992\underbrace{5988\ldots 5988}_{65\ \text{times}}$ also can be given as an answer, indeed a better one, as it is much smaller than the first suggestion.