$101$ numbers are written on a blackboard: $1^2, 2^2, 3^2, \cdots, 101^2$. Alex choses any two numbers and replaces them by their positive difference. He repeats this operation until one number is left on the blackboard. Determine the smallest possible value of this number.
1. **Initial Setup**: We start with the numbers \(1^2, 2^2, 3^2, \ldots, 101^2\) on the blackboard. The goal is to determine the smallest possible value of the final number left on the board after repeatedly replacing any two numbers with their positive difference.

2. **Sum of Squares**: The sum of the squares of the first \(101\) natural numbers is given by:
   \[
   \sum_{k=1}^{101} k^2 = \frac{101 \cdot 102 \cdot 203}{6} = 348551
   \]
   This sum is odd.

3. **Invariance of Parity**: Notice that the parity (odd or even nature) of the sum of the numbers on the board is invariant under the operation of replacing two numbers with their positive difference. This is because the difference of two numbers with the same parity is even, and the difference of two numbers with different parities is odd. Since the initial sum is odd, the final number must also be odd.

4. **Positive Difference**: Each operation reduces the number of elements on the board by one. The smallest possible value of the final number must be odd and positive.

5. **Achieving the Smallest Value**: We need to show that the smallest odd positive number, which is \(1\), can be achieved. Consider the following steps:
   - Pair numbers to form differences that eventually reduce to \(1\).
   - For example, consider the numbers \(1^2, 2^2, 3^2, \ldots, 8^2\):
     \[
     1, 4, 9, 16, 25, 36, 49, 64
     \]
     - Pair \(64\) and \(49\) to get \(15\).
     - Pair \(36\) and \(25\) to get \(11\).
     - Pair \(16\) and \(9\) to get \(7\).
     - Pair \(4\) and \(1\) to get \(3\).
     - Now, pair \(15\) and \(11\) to get \(4\).
     - Pair \(7\) and \(3\) to get \(4\).
     - Finally, pair the two \(4\)'s to get \(0\).

6. **Final Reduction**: Continue this process with the remaining numbers. Eventually, we will be left with a single number. Since the sum of the squares is odd, the final number must be odd. The smallest odd positive number is \(1\).

Conclusion:
\[
\boxed{1}
\]