Four-digit number

Let $a\neq b\neq c\neq d$ be the thousand's, hundred's, ten's and unit's digits of the number respectively. Then $0\leq a, b, c, d\leq 9$ . Also, $a$ must be equal to $1$ , since at most one digit can be zero and the number must be as close to $2006$ as possible. Then from the given condition $1001a+101b+11c+2d=2006$ we get $b+c+2d$ can be $5,15,25$ or $35$ . Also, $b$ must be the largest possible and $b+c$ must be odd. For $a=1$ and $b=9$ we have $1001+909+11c+2d=2006\implies 11c+2d=96$ . So $c\leq 8$ . Since $d\leq 9$ , therefore the only possible values of $c$ and $d$ are $8$ and $4$ respectively. Therefore the required number is $\boxed {1984}$ .

Let $N=\overline{ABCD}$ , where $A$ , $B$ , $C$ , and $D$ are single digit integer from $1$ to $9$ . Then we have:

$\begin{aligned} \overline{ABCD} + A + B + C + D & = 2006 \\ 1001A + 101B + 11C + 2D & = 2006 & \small \blue{\text{Since }\overline{A00A}+\overline{B0B} < 2006 \implies A=1, B=9} \\ 1910 + 11C + 2D & = 2006 \\ 11C + 2D & = 96 & \small \blue{\text{Putting }C=8} \\ 88 + 2D & = 96 \\ 2D & = 8 \\ \implies D & = 4 \end{aligned}$

Therefore $N = \boxed{1984}$ .