Four digit number M

Let $M$ be represented as $1000A + X$, where $A$ is the leftmost non-zero digit ($1 \le A \le 9$) and $X$ is the three-digit number formed by the remaining digits ($100 \le X \le 999$).

According to the question: $X = \frac{1}{9} M \implies 9X = 1000A + X$ $8X = 1000A \implies X = 125A$

Since $X$ is a 3-digit number: $100 \le 125A \le 999$

For $A = 1, 2, 3, 4, 5, 6, 7$, $X$ lies between 125 and 875, all of which are 3-digit numbers. For $A = 8$, $X = 1000$ (a 4-digit number, invalid).

Hence, there are 7 possible values of $M$.