Solution:

As in the figure, we number the small triangles from $T_{1}$ to $T_{9}$ and color red the triangles $T_{1}, T_{2}, T_{4}, T_{5}, T_{7}, T_{9}$, leaving the other 3 white. Letting $m_{i}$ be the number written in the triangle $T_{i}$ at a certain moment, we show that the sum over the white triangles is equal to that over the red ones, that is
$$m_{1}+m_{2}+m_{4}+m_{5}+m_{7}+m_{9}=m_{3}+m_{6}+m_{8}.$$

Certainly equality (1) holds at the beginning of the game, before Marco makes any move, since both sides of (1) equal 0. We show that, at every move, the equation remains true. Whichever 2 neighboring small triangles are chosen, one is white and the other is red; therefore only one $m_{i}$ on the right and only one $m_{j}$ on the left of equation (1) are modified. Notice, however, that both are either incremented or decremented by 1, so if the equality was satisfied before performing the move it will also be satisfied after applying it.

Suppose now that
$\left\{m_{1}, \ldots, m_{9}\right\}=\{n, \ldots, n+8\}$. Whichever 3 integers among these are chosen, their sum cannot exceed $(n+8)+(n+7)+(n+6)$, from which
$$m_{3}+m_{6}+m_{8} \leq 3 n+21$$
similarly, choosing 6 integers in the set $\{n, \ldots, n+8\}$, their sum is at least equal to the sum of the 6 smallest, hence
$$m_{1}+m_{2}+m_{4}+m_{5}+m_{7}+m_{9} \geq 6 n+15.$$
Combining the last two inequalities with (1) we obtain
$$6 n+15 \leq m_{1}+m_{2}+m_{4}+m_{5}+m_{7}+m_{9}=m_{3}+m_{6}+m_{8} \leq 3 n+21,$$
which implies $3 n \leq 6$, that is $n \leq 2$. It remains to exclude that $n=1$ can occur. In that case $m_{1}+\cdots+m_{9}=1+\cdots+9=45$, which is an odd number. However, every move of Marco changes the sum $m_{1}+\cdots+m_{9}$ by +2 or -2 (depending on whether he chooses to add or subtract from the two neighboring small triangles); since this sum equals 0 at the beginning of the game, it always remains even (and in particular different from 45), so the case $n=1$ is impossible.