Determine all integers $n\geq 1$ for which the numbers $1,2,\ldots,n$ may be (re)ordered as $a_1,a_2,\ldots,a_n$ in such a way that the average $\dfrac {a_1+a_2+\cdots + a_k} {k}$ is an integer for all values $1\leq k\leq n$.

(Dan Schwarz)
1. First, we need to ensure that the average of the first \( k \) terms is an integer for all \( 1 \leq k \leq n \). This means that for any \( k \), the sum \( a_1 + a_2 + \cdots + a_k \) must be divisible by \( k \).

2. Consider the sum of all \( n \) terms:
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{1 + 2 + \cdots + n}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}
   \]
   For this to be an integer, \( n+1 \) must be even, implying \( n \) must be odd. Therefore, \( n \equiv 1 \pmod{2} \).

3. Let's check the smallest odd \( n \), which is \( n = 1 \):
   \[
   \text{For } n = 1, \text{ the sequence is } \{1\}, \text{ and the average is } 1, \text{ which is an integer.}
   \]
   Thus, \( n = 1 \) is a solution.

4. Now, consider \( n \geq 3 \). We need to check if the sequence can be reordered such that the average of the first \( k \) terms is an integer for all \( 1 \leq k \leq n \).

5. For \( k = n-1 \):
   \[
   \frac{a_1 + a_2 + \cdots + a_{n-1}}{n-1} = \frac{\frac{n(n+1)}{2} - a_n}{n-1}
   \]
   Let \( \frac{n(n+1)/2 - a_n}{n-1} = m \) for some integer \( m \). This implies:
   \[
   n(n+1) - 2a_n = 2m(n-1)
   \]
   Since \( n \) is odd, let \( n = 2l + 1 \) for some integer \( l \). Then:
   \[
   (2l+1)(2l+2) - 2a_n = 2m(2l)
   \]
   Simplifying, we get:
   \[
   2l^2 + 3l + 1 - a_n = ml
   \]
   This implies:
   \[
   a_n \equiv 1 \pmod{l}
   \]
   Therefore, \( a_n \in \{1, \frac{n+1}{2}, n\} \).

6. Check the possible values of \( a_n \):
   - If \( a_n = 1 \):
     \[
     2l^2 + 3l = 2ml \implies 2l + 3 = 2m \implies m = l + \frac{3}{2}
     \]
     This is a contradiction since \( m \) must be an integer.
   - If \( a_n = n = 2l + 1 \):
     \[
     2l^2 + 3l + 1 - (2l + 1) = 2ml \implies 2l^2 + l = 2ml \implies l = m
     \]
     This is a contradiction since \( l \) must be an integer.
   - Therefore, \( a_n = \frac{n+1}{2} \).

7. For \( n \geq 5 \):
   \[
   \frac{a_1 + a_2 + \cdots + a_{n-2}}{n-2} = \frac{\frac{n(n+1)}{2} - a_n - a_{n-1}}{n-2}
   \]
   Substituting \( a_n = \frac{n+1}{2} \):
   \[
   \frac{\frac{n(n+1)}{2} - \frac{n+1}{2} - a_{n-1}}{n-2} = \frac{\frac{(n+1)(n-1)}{2} - a_{n-1}}{n-2}
   \]
   This implies:
   \[
   \frac{(n+1)(n-1)}{2} \equiv a_{n-1} \pmod{n-2}
   \]
   Since \( n \geq 5 \), \( \frac{n+1}{2} \leq n-2 \) and \( \frac{n+1}{2} + (n-2) > n \), forcing \( a_{n-1} = \frac{n+1}{2} = a_n \), which is a contradiction.

8. Therefore, the only possible values of \( n \) are \( n = 1 \) and \( n = 3 \).

The final answer is \( \boxed{ n \in \{1, 3\} } \).