If the length of the string is at most 26, we can build a string using each letter only once. If the length is more than 26, we should use all the letters and always put all occurrences of a letter next to each other. The answer can be calculated using dynamic programming.

For L <= 26:

The answer is 26*..*(26-L+1), since we have 26 letters for the first choice, 25 for the second one, etc. The minimum weight of a word is 0.

For L > 26:

First of all, we should use each letter at least once (since we could easily lower the weight using the unused letter). What to do with the remaining L — 26 choices? Actually, it doesn't matter which letter we pick for each of these choices, since, as long as all occurrences of a letter are next to each other, they contribute to weight identically. For example, if we already have AAB (for simplicity assume that A and B are the only letters in the alphabet) picking A and forming AAAB, or picking B and forming AABB adds the same weight (+1). So we have L — 26 "balls", and we need to distribute them to 26 "urns". The number of ways to do that is
C[balls+urns-1][urns-1] = C[L-26+26-1][26-1] = C[L-1][25]. Also, we can arbitrarily rearrange groups of identical letters, and the answer becomes
C[L-1][25] * 26!. The minimum weight of a word is L — 26.