Find all integers $n\geq 1$ such that there exists a permutation $(a_1,a_2,...,a_n)$ of $(1,2,...,n)$ such that $a_1+a_2+...+a_k$ is divisible by $k$ for $k=1,2,...,n$
1. We start by considering the condition that there exists a permutation \((a_1, a_2, \ldots, a_n)\) of \((1, 2, \ldots, n)\) such that the sum \(a_1 + a_2 + \ldots + a_k\) is divisible by \(k\) for \(k = 1, 2, \ldots, n\).

2. For \(k = n\), we have:
   \[
   a_1 + a_2 + \ldots + a_n = \frac{n(n+1)}{2}
   \]
   Since \(\frac{n(n+1)}{2}\) must be divisible by \(n\), \(n\) must be odd (because \(n\) and \(n+1\) are consecutive integers, one of which is even, making their product even, and thus \(\frac{n(n+1)}{2}\) is an integer).

3. Consider \(k = n-1\):
   \[
   a_1 + a_2 + \ldots + a_{n-1} = \frac{n(n+1)}{2} - a_n
   \]
   Since \(a_1 + a_2 + \ldots + a_{n-1}\) must be divisible by \(n-1\), we have:
   \[
   n-1 \mid \frac{n(n+1)}{2} - a_n
   \]
   Simplifying, we get:
   \[
   n-1 \mid n(n+1) - 2a_n
   \]
   \[
   n-1 \mid n^2 + n - 2a_n
   \]
   \[
   n-1 \mid (n^2 + n - 2a_n) - (n-1)^2
   \]
   \[
   n-1 \mid n^2 + n - 2a_n - (n^2 - 2n + 1)
   \]
   \[
   n-1 \mid 3n - 2a_n - 1
   \]
   \[
   n-1 \mid 2a_n - 2
   \]
   Since \(a_n \leq n\), we have \(2a_n - 2 \leq 2(n-1)\). Therefore, \(2a_n - 2\) must be either \(n-1\) or \(2(n-1)\).

4. If \(2a_n - 2 = n-1\), then:
   \[
   2a_n = n + 1
   \]
   \[
   a_n = \frac{n+1}{2}
   \]
   This implies \(n\) is odd.

5. If \(2a_n - 2 = 2(n-1)\), then:
   \[
   2a_n - 2 = 2n - 2
   \]
   \[
   2a_n = 2n
   \]
   \[
   a_n = n
   \]
   This case does not work because \(a_n\) must be a permutation of \((1, 2, \ldots, n)\).

6. Therefore, \(a_n = \frac{n+1}{2}\) and \(n\) must be odd.

7. Now consider \(k = n-2\):
   \[
   a_1 + a_2 + \ldots + a_{n-2} = \frac{n(n+1)}{2} - a_n - a_{n-1}
   \]
   \[
   n-2 \mid \frac{n(n+1)}{2} - \left(\frac{n+1}{2}\right) - a_{n-1}
   \]
   Simplifying, we get:
   \[
   n-2 \mid n^2 - 1 - 2a_{n-1}
   \]
   \[
   n-2 \mid n^2 - 1 - 2a_{n-1} - (n-1)^2 - 4(n-2)
   \]
   \[
   n-2 \mid 2a_{n-1} - 3
   \]
   Since \(2a_{n-1} - 3 \leq 2n - 3 < 3(n-2)\) for \(n > 3\), we have two cases:
   \[
   2a_{n-1} - 3 = n-2 \quad \text{or} \quad 2a_{n-1} - 3 = 2(n-2)
   \]

8. If \(2a_{n-1} - 3 = n-2\), then:
   \[
   2a_{n-1} = n + 1
   \]
   \[
   a_{n-1} = \frac{n+1}{2} = a_n
   \]
   This is a contradiction because \(a_{n-1}\) and \(a_n\) must be distinct.

9. If \(2a_{n-1} - 3 = 2(n-2)\), then:
   \[
   2a_{n-1} - 3 = 2n - 4
   \]
   \[
   2a_{n-1} = 2n - 1
   \]
   This is a contradiction because \(2a_{n-1}\) must be an integer.

10. Hence, \(n \leq 3\). Checking the values:
    - \(n = 1\) works trivially.
    - \(n = 2\) does not work because \(n\) must be odd.
    - \(n = 3\) works with permutations \((1, 3, 2)\) or \((3, 1, 2)\).

Conclusion:
The only values of \(n\) that satisfy the conditions are \(n = 1\) and \(n = 3\).

The final answer is \( \boxed{ n = 1 } \) or \(n = 3\).