Let $n$ be a integer $1<n<2010$, where we have a polygon with $2010$ sides and $n$ coins, we have to paint the vertices of this polygon with $n$ colors and we've to put the $n$ coins in $n$ vertices of the polygon. In each second the coins will go to the neighbour vertex, going in the clockwise.
Determine the values of $n$ such that is possible paint and choose the initial position of the coins where in each second the $n$ coins are in vertices of distinct colors
1. **Divisibility Condition**:
   - If \( n \) divides 2010, then it is possible to paint the vertices and place the coins such that each coin is on a vertex of a distinct color at every second.
   - For example, if \( n = 3 \), we can color the vertices in a repeating pattern: red, green, blue, red, green, blue, and so on. Place the coins on three consecutive vertices initially. As the coins move clockwise, they will always land on vertices of different colors.

2. **Non-Divisibility Condition**:
   - If \( n \) does not divide 2010, we need to show that it is impossible to satisfy the problem's requirements.
   - Consider the first 2010 seconds (one full rotation around the polygon). Suppose it is possible to color the vertices with \( n \) colors and place \( n \) coins such that each coin is on a vertex of a distinct color at every second.

3. **Color Frequency**:
   - Let \( f(k) \) be the number of vertices colored with color \( k \). Then, we have:
     \[
     f(1) + f(2) + \cdots + f(n) = 2010
     \]
   - Without loss of generality, assume \( f(1) \geq f(2) \geq \cdots \geq f(n) \). Since \( n \) does not divide 2010, there must be at least one color \( k \) such that \( f(k) > \frac{2010}{n} \).

4. **Coin Movement Analysis**:
   - Consider a coin \( i \). This coin will pass through vertices of color 1 exactly \( f(1) \) times during the first 2010 seconds.
   - Summing over all \( n \) coins, the total number of times vertices of color 1 are passed by a coin is \( n \cdot f(1) \).

5. **Contradiction**:
   - Since \( f(1) > \frac{2010}{n} \), we have:
     \[
     n \cdot f(1) > n \cdot \frac{2010}{n} = 2010
     \]
   - This implies that the total number of times vertices of color 1 are passed by a coin is greater than 2010.
   - However, since there are only 2010 vertices and each second a coin can only pass through one vertex, the total number of times vertices of color 1 are passed by a coin cannot exceed 2010. This is a contradiction.

Therefore, if \( n \) does not divide 2010, it is impossible to paint the vertices and place the coins as required.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) divides 2010.