Find the largest value of n, such that there exists a polygon with n sides, 2 adjacent sides of length 1, and all his diagonals have an integer length.
1. Let \( V_1, V_2, V_3, \ldots, V_n \) be the vertices of the polygon. Without loss of generality (WLOG), assume \( V_1V_2 = V_2V_3 = 1 \). By the triangle inequality on \( \triangle V_1V_2V_3 \), we have:
   \[
   V_1V_3 < V_1V_2 + V_2V_3 = 1 + 1 = 2
   \]
   Since \( V_1V_3 \) must be an integer, the only possible value is \( V_1V_3 = 1 \). Thus, \( \triangle V_1V_2V_3 \) is an equilateral triangle with side length 1.

2. Suppose \( n \geq 4 \). Consider the diagonal \( V_1V_4 \). By the triangle inequality on \( \triangle V_1V_2V_4 \), we have:
   \[
   V_1V_4 < V_1V_2 + V_2V_4 = 1 + V_2V_4
   \]
   Similarly, by the triangle inequality on \( \triangle V_2V_3V_4 \), we have:
   \[
   V_2V_4 < V_2V_3 + V_3V_4 = 1 + V_3V_4
   \]
   Since \( V_1V_4 \) and \( V_2V_4 \) must be integers, and \( V_1V_4 = V_2V_4 \), we conclude that \( V_1V_4 = V_2V_4 \).

3. Now, consider \( n \geq 6 \). We have \( V_1V_4 = V_2V_4 \) and \( V_1V_5 = V_2V_5 \). However, \( V_3, V_4, \) and \( V_5 \) are on the perpendicular bisector of \( V_1V_2 \), which leads to a contradiction. Therefore, \( n \leq 5 \).

4. For \( n = 5 \), consider a concave polygon. If we add the convex condition, the maximum value of \( n \) is 4.

The final answer is \( \boxed{ n = 4 } \) for a convex polygon and \( n = 5 \) for a concave polygon.