Colin has $900$ Choco Pies. He realizes that for some integer values of $n \le 900$, if he eats n pies a day, he will be able to eat the same number of pies every day until he runs out. How many possible values of $n$ are there?
To determine how many possible values of \( n \) there are such that Colin can eat \( n \) pies a day and finish exactly in an integer number of days, we need to find the number of divisors of 900. 

1. **Prime Factorization of 900**:
   \[
   900 = 2^2 \cdot 3^2 \cdot 5^2
   \]

2. **Using the formula for the number of divisors**:
   If a number \( N \) has a prime factorization of the form:
   \[
   N = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_k^{e_k}
   \]
   then the number of divisors \( d(N) \) is given by:
   \[
   d(N) = (e_1 + 1)(e_2 + 1) \ldots (e_k + 1)
   \]

3. **Applying the formula to 900**:
   For \( 900 = 2^2 \cdot 3^2 \cdot 5^2 \):
   \[
   d(900) = (2+1)(2+1)(2+1) = 3 \cdot 3 \cdot 3 = 27
   \]

Thus, there are 27 possible values of \( n \) such that Colin can eat \( n \) pies a day and finish exactly in an integer number of days.

The final answer is \(\boxed{27}\).