Find all the finite sequences with $ n$ consecutive natural numbers $ a_1, a_2,..., a_n$, with $ n\geq3$ such that $ a_1\plus{} a_2\plus{}...\plus{} a_n\equal{}2009$.
1. Let's assume that \( a_1, a_2, \ldots, a_n \) are consecutive natural numbers. Let \( a_1 = a \). Then the sequence can be written as \( a, a+1, a+2, \ldots, a+n-1 \).

2. The sum of the first \( n \) consecutive natural numbers starting from \( a \) is given by:
   \[
   a + (a+1) + (a+2) + \cdots + (a+n-1)
   \]
   This can be simplified using the formula for the sum of an arithmetic series:
   \[
   S = \frac{n}{2} \left(2a + (n-1)\right)
   \]
   Given that this sum equals 2009, we have:
   \[
   \frac{n}{2} \left(2a + (n-1)\right) = 2009
   \]

3. To find \( n \) and \( a \), we need to solve the equation:
   \[
   n(2a + n - 1) = 4018
   \]
   This equation must be satisfied by natural numbers \( n \geq 3 \) and \( a \).

4. We will test different values of \( n \) to see if we can find a solution where \( a \) is a natural number.

   - For \( n = 3 \):
     \[
     3(2a + 3 - 1) = 4018 \implies 3(2a + 2) = 4018 \implies 6a + 6 = 4018 \implies 6a = 4012 \implies a = \frac{4012}{6} \approx 668.67 \quad (\text{not an integer})
     \]

   - For \( n = 4 \):
     \[
     4(2a + 4 - 1) = 4018 \implies 4(2a + 3) = 4018 \implies 8a + 12 = 4018 \implies 8a = 4006 \implies a = \frac{4006}{8} \approx 500.75 \quad (\text{not an integer})
     \]

   - For \( n = 5 \):
     \[
     5(2a + 5 - 1) = 4018 \implies 5(2a + 4) = 4018 \implies 10a + 20 = 4018 \implies 10a = 3998 \implies a = \frac{3998}{10} = 399.8 \quad (\text{not an integer})
     \]

   - For \( n = 6 \):
     \[
     6(2a + 6 - 1) = 4018 \implies 6(2a + 5) = 4018 \implies 12a + 30 = 4018 \implies 12a = 3988 \implies a = \frac{3988}{12} \approx 332.33 \quad (\text{not an integer})
     \]

   - For \( n = 7 \):
     \[
     7(2a + 7 - 1) = 4018 \implies 7(2a + 6) = 4018 \implies 14a + 42 = 4018 \implies 14a = 3976 \implies a = \frac{3976}{14} \approx 284 \quad (\text{integer})
     \]
     So, \( n = 7 \) and \( a = 284 \).

5. Verify the solution:
   \[
   a_1 = 284, \quad a_2 = 285, \quad a_3 = 286, \quad a_4 = 287, \quad a_5 = 288, \quad a_6 = 289, \quad a_7 = 290
   \]
   Sum:
   \[
   284 + 285 + 286 + 287 + 288 + 289 + 290 = 2009
   \]
   This confirms that the sequence is correct.

Conclusion:
The only sequence of \( n \geq 3 \) consecutive natural numbers that sums to 2009 is \( 284, 285, 286, 287, 288, 289, 290 \).

The final answer is \( \boxed{ 284, 285, 286, 287, 288, 289, 290 } \).