Label sides of a regular $n$-gon in clockwise direction in order 1,2,..,n. Determine all integers n ($n\geq 4$) satisfying the following conditions:
(1) $n-3$ non-intersecting diagonals in the $n$-gon are selected, which subdivide the $n$-gon into $n-2$ non-overlapping triangles;
(2) each of the chosen $n-3$ diagonals are labeled with an integer, such that the sum of labeled numbers on three sides of each triangles in (1) is equal to the others;
1. **Understanding the Problem:**
   We need to find all integers \( n \) (where \( n \geq 4 \)) such that:
   - We can select \( n-3 \) non-intersecting diagonals in an \( n \)-gon to subdivide it into \( n-2 \) non-overlapping triangles.
   - Each of the chosen \( n-3 \) diagonals is labeled with an integer, and the sum of the labeled numbers on the three sides of each triangle is the same for all triangles.

2. **Triangulation of the \( n \)-gon:**
   By selecting \( n-3 \) non-intersecting diagonals, we can subdivide the \( n \)-gon into \( n-2 \) triangles. This is a well-known result in combinatorial geometry and can be achieved by any triangulation of the \( n \)-gon.

3. **Labeling the Sides:**
   We label the sides of the \( n \)-gon in clockwise order as \( 1, 2, \ldots, n \). We also label the \( n-3 \) diagonals with integers such that the sum of the labeled numbers on the three sides of each triangle is the same.

4. **Sum of Labels:**
   Let the sum of the labels on the three sides of each triangle be \( S \). Since there are \( n-2 \) triangles, and each side of the \( n \)-gon and each diagonal is counted exactly twice (once for each triangle it belongs to), we have:
   \[
   2S = \sum_{i=1}^{n} i + \sum_{j=1}^{n-3} d_j
   \]
   where \( d_j \) are the labels of the diagonals.

5. **Sum of the First \( n \) Natural Numbers:**
   The sum of the first \( n \) natural numbers is:
   \[
   \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
   \]

6. **Condition for Equal Sums:**
   For the sum \( S \) to be the same for all triangles, the total sum of the labels must be divisible by \( n-2 \):
   \[
   2S = \frac{n(n+1)}{2} + \sum_{j=1}^{n-3} d_j
   \]
   \[
   S = \frac{1}{2} \left( \frac{n(n+1)}{2} + \sum_{j=1}^{n-3} d_j \right)
   \]

7. **Checking for Even \( n \):**
   For even \( n \), say \( n = 2k \):
   \[
   \sum_{i=1}^{2k} i = k(2k+1)
   \]
   This sum is always even, and thus can be divided by \( n-2 = 2k-2 \).

8. **Checking for \( n = 4k+1 \):**
   For \( n = 4k+1 \):
   \[
   \sum_{i=1}^{4k+1} i = \frac{(4k+1)(4k+2)}{2} = (4k+1)(2k+1)
   \]
   This sum is also even, and thus can be divided by \( n-2 = 4k-1 \).

9. **Conclusion:**
   From the above analysis, we can conclude that the conditions are satisfied for all even \( n \) and for \( n = 4k+1 \).

The final answer is \( \boxed{ n } \) is even or \( n = 4k+1 \).