I think it's enough to take 56 strings. Use pigeonhole principle where pigeons are strings and holes are ordered pairs (m, n) such that m + n ≤ 9.
Let s_{1}, s_{2}, ..., s_{101} be 101 bit strings of length at most 9. Prove that there exist two strings, s_{i} and s_{j}, where i not equal to j, that contain the same number of 0s and the same number of 1s. (eg: strings 001001 and 101000 contain the same number of 0s and the same number of 1s)
Below, the variables m, n and k range over nonnegative integers. So, the set of holes is H = {(m, n) | m + n ≤ 9}. Each string is mapped into a hole from H: if s is a string, m is the number of 0's in s and n is the number of 1's in s, then m + n ≤ 9 and s is mapped into (m, n). If we prove that the number of strings is strictly greater than the number of holes, then there must be at least two strings in one hole, which is what the problem asks to prove.
Note that H = H_{0} ∪ H_{1} ∪ ... ∪ H_{9} where ∪ denotes the union of sets and H_{k} = {(m, n) | m + n = k} for each 0 ≤ k ≤ 9. Also, all H_{k} are disjoint. Therefore, |H| (i.e., the number of elements in H) equals |H_{0}| + |H_{1}| + ... + |H_{9}|. Find |H_{k}| and |H| and compare it with 101.