A prison has $2004$ cells, numbered $1$ through $2004$. A jailer, carrying out the terms of a partial amnesty, unlocked every cell. Next he locked every second cell. Then he turned the key in every third cell, locking the opened cells, and unlocking the locked ones. He continued this way, on $n^{\text{th}}$ trip, turning the key in every $n^{\text{th}}$ cell, and he finished his mission after $2004$ trips. How many prisoners were released?
1. **Understanding the Problem:**
   - The jailer performs a sequence of operations on 2004 cells.
   - Initially, all cells are unlocked.
   - On the $n^{\text{th}}$ trip, the jailer toggles (locks if unlocked, unlocks if locked) every $n^{\text{th}}$ cell.
   - The jailer makes 2004 trips in total.

2. **Analyzing the Cell Toggles:**
   - A cell numbered $k$ will be toggled once for each of its divisors.
   - For example, cell 12 will be toggled on trips 1, 2, 3, 4, 6, and 12.
   - Generally, a number $k$ has an even number of divisors unless it is a perfect square. This is because divisors usually come in pairs $(d, \frac{k}{d})$.

3. **Perfect Squares:**
   - A perfect square has an odd number of divisors because one of the divisors is repeated (e.g., for $k = 36$, the divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36, where 6 is repeated).
   - Therefore, only cells numbered with perfect squares will be toggled an odd number of times, ending up unlocked.

4. **Counting Perfect Squares:**
   - We need to count the perfect squares between 1 and 2004.
   - The largest integer $n$ such that $n^2 \leq 2004$ is $\lfloor \sqrt{2004} \rfloor$.
   - Calculating $\sqrt{2004}$:
     \[
     \sqrt{2004} \approx 44.8
     \]
   - Thus, $\lfloor \sqrt{2004} \rfloor = 44$.

5. **Conclusion:**
   - There are 44 perfect squares between 1 and 2004.
   - Therefore, 44 cells will remain unlocked.

The final answer is $\boxed{44}$