Answer: 109.

Solution. Let $k \geq 100$. Suppose that for any 10 numbers of one color, their sum is either a number of the other color or does not belong to M. Let the number 1 be of the first color, and the number 2 be of the second color. Then the numbers 10 and 20 are of the second and first colors, respectively. Next, the number $28=10+9 \cdot 2$ is of the first color, and $29=20+9 \cdot 1$ is of the second color, $47=29+9 \cdot 2$ is of the first color, and $37=28+9 \cdot 1$ is of the second color. If 3 is of the first color, then $20+9 \cdot 3=47$ is of the second color. Contradiction. If 3 is of the second color, then $10+9 \cdot 3=37$ is of the first color. Contradiction. Let the numbers $1,2, \ldots, k$ be numbers of the first color $(9 \geq k \geq 2)$. Then the numbers $10,11, \ldots, 10 k$ are of the second color, and the numbers $100,101, \ldots 100 k$ are of the first color. Then the number $100+$ $9 \cdot 1=109$ should be a number of the second color. Contradiction. We will show that there is a coloring of the numbers such that for $k=108$ the specified 10 numbers will not exist. Color the numbers $1,2, \ldots, 9$ in the first color, the numbers 10, 11, .., 99 in the second color, and the numbers 100, 101, .., 108 in the first color. The sum of any 10 numbers of one color is either a number of the other color or is not a number from the set M.

Comments. Only the correct answer - $\underline{0 \text { points. A correct example with } k=109-\underline{3 \text { points }} .}$
10.4 Find the smallest positive integer $k$, for which in any coloring of the numbers of the set $M=\{1,2,3, \ldots, k\}$ in two colors, there will be ten not necessarily distinct numbers of the same color from the set $M$