Some language has only three letters - $A, B$ and $C$. A sequence of letters is called a word iff it contains exactly 100 letters such that exactly 40 of them are consonants and other 60 letters are all $A$. What is the maximum numbers of words one can pick such that any two picked words have at least one position where they both have consonants, but different consonants?
1. **Problem Restatement and Setup:**
   We are given a language with three letters: \(A\), \(B\), and \(C\). A word in this language is defined as a sequence of exactly 100 letters, with exactly 40 consonants (either \(B\) or \(C\)) and 60 vowels (\(A\)). We need to determine the maximum number of such words that can be picked such that any two picked words have at least one position where they both have consonants, but different consonants.

2. **Construction of Words:**
   Consider the 40 positions that will be occupied by consonants. Each of these positions can be either \(B\) or \(C\). Therefore, there are \(2^{40}\) possible sequences of consonants. For each of these sequences, we can append 60 \(A\)'s to form a word of 100 letters. This gives us \(2^{40}\) distinct words.

3. **Verification of the Condition:**
   We need to verify that any two distinct words from this set have at least one position where they both have consonants, but different consonants. Consider two distinct sequences of consonants. Since they are distinct, there is at least one position where they differ. At this position, one word has \(B\) and the other has \(C\), satisfying the condition.

4. **Inductive Argument:**
   To generalize, we can use induction on the number of \(A\)'s. For \(k = 0\), the problem is trivial as there are no \(A\)'s, and the maximum number of words is \(2^{40}\). Assume the result holds for \(k\) \(A\)'s. Now consider \(k+1\) \(A\)'s. Partition the words into sets based on the last letter being \(A\), \(B\), or \(C\). For the sets ending in \(B\) or \(C\), change the last letter to \(A\) and consider the resulting sets. By the inductive hypothesis, each of these sets has at most \(2^{40}\) words. Summing these and dividing by 2, we get the desired bound.

5. **Conclusion:**
   By the construction and verification, we have shown that the maximum number of words is \(2^{40}\).

The final answer is \(\boxed{2^{40}}\).