7. Label the $n$ sides of a regular $n(n \geqslant 4)$-sided polygon in a clockwise direction with $1,2, \cdots, n$. Find all integers $n$ such that the $n$-sided polygon can be divided into $n-2$ triangular regions using $n-3$ non-intersecting diagonals inside the polygon, and each of these $n-3$ diagonals is labeled with an integer, satisfying that the sum of the numbers on the three sides of each triangle is equal.
7. First, prove: if $n$ satisfies the condition, then $n \not \equiv 2(\bmod 4)$.
If a positive integer $n$ satisfies the condition, let the sum of the integers marked on the three sides of each triangle be $S$, and the sum of the integers marked on all $n-3$ diagonals be $m$. Summing the sums of the numbers marked on the three sides of all $n-2$ triangles, each diagonal is counted twice, we get
$$
(n-2) S=(1+2+\cdots+n)+2 m .
$$

Assume $n \equiv 2(\bmod 4)$. Then $(n-2) S$ is even.
However, $(1+2+\cdots+n)+2 m$ is odd, which is a contradiction.
Therefore, $n \neq 2(\bmod 4)$.
Next, use mathematical induction to prove: all $n \neq \equiv$ $2(\bmod 4)$ satisfy the condition.

As shown in Figure 4, constructions that satisfy the condition are given for $n=4,5,7$.

It is only necessary to prove: if $n$ satisfies the condition, then $n+4$ also satisfies the condition.

As shown in Figure 5, first divide the convex $n+4$-gon into a convex $n$-gon and a convex hexagon using one diagonal, and mark this diagonal with $n$.

Since $n$ satisfies the condition, the convex $n$-gon can be divided into triangles such that the sum of the numbers marked on the three sides of each triangle is equal (let this sum be $S$). Divide the convex hexagon into four triangles using three diagonals as shown in Figure 5, and mark the three diagonals with $S-2 n-1$, $n+1$, and $S-2 n-5$ respectively. It can be verified that the sum of the numbers marked on the three sides of these four triangles also equals $S$. Therefore, $n+4$ also satisfies the condition.
In summary, the $n$ that satisfy the condition are $n \neq 2(\bmod 4)$.