Without loss of generality, assume the large equilateral triangle points upward. We will prove the following: among all the small triangles, the number of upward-pointing equilateral triangles exceeds the number of downward-pointing equilateral triangles by at least $n$. This naturally proves the original statement.

If all the small triangles are equilateral triangles, this obviously holds (since in each horizontal row, the number of upward-pointing triangles exceeds the number of downward-pointing ones by exactly one). If not, then perform the following operation: find the longest edge $AB$ among all the small triangles, and consider the two small triangles $ABC$ and $ABD$ having this edge. We can prove the following:

Lemma 1: $ABCD$ forms a parallelogram.
Proof: By Pick's formula, there are no other grid points on the segment $AB$, and all possible positions for $C$ and $D$ lie on two lines parallel to $AB$ (one on each side of $AB$), spaced at intervals equal to the length of $AB$. Moreover, since $AB$ is the longest edge, the angles $\angle CBA$ and $\angle CAB$ must be acute, so the foot of the perpendicular from $C$ to $AB$ falls within the segment $AB$. Combining these two facts, we know the position of $C$ is uniquely determined; similarly, $D$ is also uniquely determined, and by symmetry it is easy to see that $D$ is the point such that $ABCD$ forms a parallelogram. Q.E.D.

Based on the fact that these two small triangles $ABC$ and $ABD$ form a parallelogram, replacing $AB$ with $CD$ yields a new decomposition satisfying the conditions of the problem, and

Lemma 2: $AB > CD$.
Proof: Since $AB \ge \sqrt{3}$, we know $d(C, AB) \le 1/2$; and since $AC \ge 1$, we know $\angle CAB \le 30^\circ$. Similarly, $\angle CBA \le 30^\circ$. Therefore $\angle C$ is obtuse, so by the law of cosines (pivot theorem), $AB > CD$. Q.E.D.

Therefore, each time we perform this operation, the length of the longest edge decreases; noting that there are only finitely many possible distances between two points on the grid, we can repeat this operation until the longest edge is reduced to length $1$, that is, until the state in which all small triangles are equilateral triangles is reached. Moreover, in each such operation, it is easy to see that the difference between the number of upward-pointing and downward-pointing equilateral triangles does not change. Hence we know that in the original decomposition, the number of upward-pointing equilateral triangles exceeds the number of downward-pointing small equilateral triangles by $n$. Q.E.D.