, 2014

Let $N_0 = 0$. For $1 \le i \le k$, let $N_i$ be the number of babies born on or before Day $i$, and let $n_i = N_i - N_{i-1}$ be the number of babies born on Day $i$. Let $S = \{N_1, \dots, N_k\}$ and $T = \{N_1 + 100, \dots, N_k + 100\}$. Because at least one baby is born each day, both sets contain $k$ distinct integers between 1 and 2114, inclusive. The sets $S$ and $T$ might intersect: in fact, they intersect if and only if there are a pair of indices $i$ and $j$ with $N_j = N_i + 100$ for some $i < j$, which is equivalent to the number of babies born in the period between days $i+1$ and $j$ inclusive being 100.
If there were at least 12 different integers in $S$ having the same remainder mod 100, there would also be at least 12 different integers in $T$ with this remainder. But between 1 and 2114, no remainder mod 100 occurs more than 22 times, hence $S \cap T$ cannot be empty in this case.
Assume now that no remainder mod 100 occurs more than eleven times in $S$. If $k \ge 1015$, then there are at least 15 remainders mod 100 that occur at least eleven times in $S$, and consequently also at least eleven times in $T$. Thus $S \cup T$ has at least 15 remainders mod 100 that each occur at least 22 times (including repetitions, in the case of any remainders that occur in both $S$ and $T$). Since there are only 14 remainders that occur 22 times between 1 and 2114, and none that occur more frequently than that, some of these occurrences must overlap. Thus, $S$ and $T$ must intersect, proving (a).
Let $n_i = 1$ if $i \in N$ is not a multiple of 100, and $n_i = 101$ if $i$ is a multiple of 100. It is readily verified that this pattern ensures that $N_j - N_i \ne 100$ for all $0 \le i \le j$. Also $N_{100s+j} = 200s + j$ for all integers $0 \le j < 100$, $s \ge 0$. In particular, $N_{1014} = 2014$, as required for $k = 1014$ in part (b).