2018 little ducklings numbered 1 through 2018 are standing in a line, with each holding a slip of paper with a nonnegative number on it; it is given that ducklings 1 and 2018 have the number zero.  At some point, ducklings 2 through 2017 change their number to equal the average of the numbers of the ducklings to their left and right.  Suppose the new numbers on the ducklings sum to 1000.  What is the maximum possible sum of the original numbers on all 2018 slips?
1. Let \( d_i \) be the number that duckling \( i \) originally has, and \( d_i' \) be the number that duckling \( i \) has after changing their numbers. We are given that \( d_1 = 0 \) and \( d_{2018} = 0 \).

2. According to the problem, ducklings 2 through 2017 change their number to the average of the numbers of the ducklings to their left and right. Therefore, we have the following equations:
   \[
   d_2' = \frac{d_1 + d_3}{2} = \frac{d_3}{2}
   \]
   \[
   d_3' = \frac{d_2 + d_4}{2}
   \]
   \[
   d_4' = \frac{d_3 + d_5}{2}
   \]
   \[
   \vdots
   \]
   \[
   d_{2016}' = \frac{d_{2015} + d_{2017}}{2}
   \]
   \[
   d_{2017}' = \frac{d_{2016} + d_{2018}}{2} = \frac{d_{2016}}{2}
   \]

3. Summing these new numbers, we get:
   \[
   d_2' + d_3' + d_4' + \cdots + d_{2016}' + d_{2017}' = 1000
   \]

4. Substituting the expressions for \( d_i' \) in terms of \( d_i \), we get:
   \[
   d_2' + d_3' + d_4' + \cdots + d_{2016}' + d_{2017}' = \frac{d_3}{2} + \frac{d_2 + d_4}{2} + \frac{d_3 + d_5}{2} + \cdots + \frac{d_{2015} + d_{2017}}{2} + \frac{d_{2016}}{2}
   \]

5. Rearranging and combining like terms, we get:
   \[
   1000 = \frac{d_2}{2} + d_3 + d_4 + \cdots + d_{2016} + \frac{d_{2017}}{2}
   \]

6. Let \( S \) be the sum of the original numbers on all 2018 slips:
   \[
   S = d_1 + d_2 + d_3 + \cdots + d_{2017} + d_{2018}
   \]
   Since \( d_1 = 0 \) and \( d_{2018} = 0 \), we have:
   \[
   S = d_2 + d_3 + d_4 + \cdots + d_{2017}
   \]

7. From the equation derived in step 5, we have:
   \[
   S = 1000 + \frac{d_2}{2} + \frac{d_{2017}}{2}
   \]

8. To maximize \( S \), we need to maximize \( \frac{d_2}{2} + \frac{d_{2017}}{2} \). Since all numbers are nonnegative, the maximum value of \( \frac{d_2}{2} + \frac{d_{2017}}{2} \) is 1000. This is achieved when \( d_2 = 1000 \) and \( d_{2017} = 1000 \).

9. Therefore, the maximum possible sum of the original numbers on all 2018 slips is:
   \[
   S = 1000 + 1000 = 2000
   \]

The final answer is \(\boxed{2000}\)