A scientist begins an experiment with a cell culture that starts with some integer number of identical cells. After the first second, one of the cells dies, and every two seconds from there another cell will die (so one cell dies every odd-numbered second from the starting time). Furthermore, after exactly $60$ seconds, all of the living cells simultaneously split into two identical copies of itself, and this continues to happen every $60$ seconds thereafter. After performing the experiment for awhile, the scientist realizes the population of the culture will be unbounded and quickly shuts down the experiment before the cells take over the world. What is the smallest number of cells that the experiment could have started with?

$\text{(A) }30\qquad\text{(B) }31\qquad\text{(C) }60\qquad\text{(D) }61\qquad\text{(E) }62$
1. Let's denote the initial number of cells as \( N \).
2. We need to determine the smallest \( N \) such that the population of the culture becomes unbounded over time.
3. Cells die every odd-numbered second. Therefore, in the first 60 seconds, the number of cells that die is:
   \[
   1 + 1 + 1 + \ldots + 1 \quad \text{(30 times)}
   \]
   This is because there are 30 odd-numbered seconds in the first 60 seconds.
4. Therefore, the number of cells that die in the first 60 seconds is 30.
5. After 60 seconds, the remaining number of cells is:
   \[
   N - 30
   \]
6. At the 60-second mark, the remaining cells double. Thus, the number of cells after doubling is:
   \[
   2 \times (N - 30)
   \]
7. For the population to be unbounded, the number of cells after each 60-second interval must increase. This means:
   \[
   2 \times (N - 30) > N
   \]
8. Solving the inequality:
   \[
   2N - 60 > N
   \]
   \[
   N > 60
   \]
9. Therefore, the smallest integer \( N \) that satisfies this inequality is \( N = 61 \).

The final answer is \( \boxed{61} \)