Given a set of points in space, a [i]jump[/i] consists of taking two points, $P$ and $Q,$ and replacing $P$ with the reflection of $P$ over $Q$. Find the smallest number $n$ such that for any set of $n$ lattice points in $10$-dimensional-space, it is possible to perform a finite number of jumps so that some two points coincide.

[i]Author: Anderson Wang[/i]
1. **Base Case:**
   - Consider any 3 points in 1-dimensional space. We need to show that we can make some two points coincide.
   - Let the points be \( P_1, P_2, P_3 \) with coordinates \( x_1, x_2, x_3 \) respectively.
   - Without loss of generality, assume \( x_1 \leq x_2 \leq x_3 \).
   - Perform a jump by reflecting \( P_1 \) over \( P_2 \). The new position of \( P_1 \) will be \( 2x_2 - x_1 \).
   - The distance between the new \( P_1 \) and \( P_3 \) is \( |2x_2 - x_1 - x_3| \), which is less than the original distance \( |x_1 - x_3| \).
   - Repeating this process, we can eventually make two points coincide.

2. **Inductive Step:**
   - Assume that for any set of \( 2^n + 1 \) points in \( n \)-dimensional space, it is possible to perform a finite number of jumps so that some two points coincide.
   - We need to show that for any set of \( 2^{n+1} + 1 \) points in \( (n+1) \)-dimensional space, it is possible to perform a finite number of jumps so that some two points coincide.
   - Consider the set of \( 2^{n+1} + 1 \) points in \( (n+1) \)-dimensional space. Project these points onto a 1-dimensional space by considering only the first coordinate.
   - Divide the points into two sets \( A \) and \( B \) based on the parity of the first coordinate.
   - By the Pigeonhole Principle, at least one of the sets \( A \) or \( B \) has at least \( 2^n + 1 \) points. Assume without loss of generality that \( A \) has at least \( 2^n + 1 \) points.
   - Choose any point \( P \) from \( B \). By manipulating the points in \( A \) and \( P \), we can make two points in \( A \) coincide (using the base case and parity argument).
   - Once two points in \( A \) coincide, they lie in the same \( n \)-dimensional plane.
   - Repeat this process, treating the coinciding points as a unit, to accumulate all points in \( A \) into an \( n \)-dimensional plane.
   - By the inductive hypothesis, we can then make some two points in \( A \) coincide.

Thus, by induction, for any set of \( 2^{10} + 1 = 1025 \) points in 10-dimensional space, it is possible to perform a finite number of jumps so that some two points coincide.

The final answer is \( \boxed{ 1025 } \).