How many alphabetic sequences (that is, sequences containing only letters from $a\cdots z$) of length $2013$ have letters in alphabetic order?
1. **Understanding the problem**: We need to find the number of alphabetic sequences of length 2013 where the letters are in alphabetical order. This means that each letter in the sequence is either the same as or comes after the previous letter in the alphabet.

2. **Rephrasing the problem**: We can think of this problem as distributing 2013 identical items (letters) into 26 distinct categories (letters from 'a' to 'z') such that the order of categories is preserved. This is a classic "stars and bars" problem in combinatorics.

3. **Applying the stars and bars theorem**: The stars and bars theorem states that the number of ways to distribute \( n \) identical items into \( k \) distinct categories is given by:
   \[
   \binom{n + k - 1}{k - 1}
   \]
   Here, \( n = 2013 \) (the length of the sequence) and \( k = 26 \) (the number of letters in the alphabet).

4. **Substituting the values**: We substitute \( n = 2013 \) and \( k = 26 \) into the formula:
   \[
   \binom{2013 + 26 - 1}{26 - 1} = \binom{2038}{25}
   \]

5. **Conclusion**: The number of alphabetic sequences of length 2013 with letters in alphabetical order is given by:
   \[
   \boxed{\binom{2038}{25}}
   \]