Find all integers that can be written in the form $\frac{1}{a_1}+\frac{2}{a_2}+...+\frac{9}{a_9}$ where $a_1,a_2, ...,a_9$ are nonzero digits, not necessarily different. 
1. **Define the function \( F(a_1, a_2, \ldots, a_9) \):**
   \[
   F(a_1, a_2, \ldots, a_9) = \frac{1}{a_1} + \frac{2}{a_2} + \cdots + \frac{9}{a_9}
   \]
   where \( a_1, a_2, \ldots, a_9 \) are nonzero digits (i.e., \( 1, 2, \ldots, 9 \)).

2. **Determine the minimum and maximum possible values of \( F \):**
   - The minimum value occurs when all \( a_i = 9 \):
     \[
     F(9, 9, 9, 9, 9, 9, 9, 9, 9) = \frac{1}{9} + \frac{2}{9} + \cdots + \frac{9}{9} = \frac{1+2+\cdots+9}{9} = \frac{45}{9} = 5
     \]
   - The maximum value occurs when all \( a_i = 1 \):
     \[
     F(1, 1, 1, 1, 1, 1, 1, 1, 1) = 1 + 2 + 3 + \cdots + 9 = 45
     \]

3. **Verify intermediate values:**
   - For \( F = 6 \):
     \[
     F(9, 2, 9, 9, 9, 9, 7, 9, 9) = \frac{1}{9} + \frac{2}{2} + \frac{3}{9} + \frac{4}{9} + \frac{5}{9} + \frac{6}{9} + \frac{7}{7} + \frac{8}{9} + \frac{9}{9} = \frac{1}{9} + 1 + \frac{1}{3} + \frac{4}{9} + \frac{5}{9} + \frac{2}{3} + 1 + \frac{8}{9} + 1 = 6
     \]
   - For \( F = 7 \):
     \[
     F(2, 4, 6, 8, 5, 6, 7, 8, 9) = \frac{1}{2} + \frac{2}{4} + \frac{3}{6} + \frac{4}{8} + \frac{5}{5} + \frac{6}{6} + \frac{7}{7} + \frac{8}{8} + \frac{9}{9} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 1 + 1 + 1 + 1 + 1 = 7
     \]
   - For \( F = 8 \):
     \[
     F(1, 4, 6, 4, 5, 6, 7, 8, 9) = \frac{1}{1} + \frac{2}{4} + \frac{3}{6} + \frac{4}{4} + \frac{5}{5} + \frac{6}{6} + \frac{7}{7} + \frac{8}{8} + \frac{9}{9} = 1 + \frac{1}{2} + \frac{1}{2} + 1 + 1 + 1 + 1 + 1 + 1 = 8
     \]
   - For \( F = 9 \):
     \[
     F(1, 2, 3, 4, 5, 6, 7, 8, 9) = \frac{1}{1} + \frac{2}{2} + \frac{3}{3} + \frac{4}{4} + \frac{5}{5} + \frac{6}{6} + \frac{7}{7} + \frac{8}{8} + \frac{9}{9} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9
     \]

4. **Algorithm to generate all integers between 5 and 45:**
   - Start with \( G = (1, 2, 3, 4, 5, 6, 7, 8, 9) \) which gives \( F(G) = 9 \).
   - Apply the following steps to increment \( F(G) \) by 1:
     1. If all \( a_k = 1 \), do nothing.
     2. Else:
        2.1. Let \( k \) be the smallest index such that \( a_k = 1 \) and \( k \neq 1 \).
        2.2. If \( k \) does not exist or \( a_{k+1} = 1 \) or \( k = 9 \):
             2.2.1. Set \( a_2 = 1 \).
        2.3. Else:
             2.3.1. Set \( a_k = k \) and \( a_{k+1} = 1 \).

   - This algorithm ensures that each step increases \( F(G) \) by 1, covering all integers from 9 to 45.

5. **Conclusion:**
   - The function \( F \) can take any integer value from 5 to 45.

The final answer is \( \boxed{ [5, 45] } \)