Consider the infinite grid of lattice points in $\mathbb{Z}^3$. Little D and Big Z play a game, where Little D first loses a shoe on an unmunched point in the grid. Then, Big Z munches a shoe-free plane perpendicular to one of the coordinate axes. They continue to alternate turns in this fashion, with Little D's goal to lose a shoe on each of $n$ consecutive lattice points on a line parallel to one of the coordinate axes. Determine all $n$ for which Little D can accomplish his goal.

[i]David Yang.[/i]
1. **Initial Setup and Definitions:**
   - Little D and Big Z play a game on an infinite grid of lattice points in \(\mathbb{Z}^3\).
   - Little D loses a shoe on an unmunched point in the grid.
   - Big Z munches a shoe-free plane perpendicular to one of the coordinate axes.
   - Little D's goal is to lose a shoe on each of \(n\) consecutive lattice points on a line parallel to one of the coordinate axes.

2. **Coloring Scheme:**
   - Define \(X = Y = Z = \mathbb{Z}\) (each representing an axis).
   - If Little D loses a shoe on \((x, y, z)\), color \(x \in X\), \(y \in Y\), and \(z \in Z\) red.
   - If Big Z munches a \(yz\)-plane with \(x\)-intercept \(x\), color \(x \in X\) blue; similarly for the other two cases.

3. **Red Intervals:**
   - A red interval is a contiguous sequence of red elements \(\{a, a+1, \dots, b\}\) in one of \(X\), \(Y\), or \(Z\).
   - The length of a red interval is the number of elements in it.
   - A red interval is alive if none of \(\{a, \dots, a+n-1\}\) are blue (meaning it could be completed on the right to a red interval of length \(n\)).

4. **Claim:**
   - For each \(\ell = 1, 2, \dots, n\), Little D can create arbitrarily many alive red intervals of length \(\ell\) on the same axis.

5. **Proof by Induction:**
   - **Base Case (\(\ell = 1\)):**
     - Drop shoes on \((n, n, n)\), \((2n, 2n, 2n)\), and so on.
     - After \(m\) shoes are dropped, there will be at least \(2m\) alive red intervals of length 1, since Big Z can only kill at most one red interval per turn.
     - By the pigeonhole principle, some axis will have at least \(\frac{2}{3}m\) alive red intervals.

   - **Inductive Step:**
     - Assume we have sufficiently many alive red intervals of length \(\ell\) on an axis, say \(X\).
     - Perform the following procedure iteratively for \(\ell+1\) turns:
       - Select a large integer \(t\).
       - On each turn, Little D drops a shoe on \((x, y, z)\) where:
         - \(x\) is chosen to extend a red interval in \(X\) by one.
         - \(y = t, t+1, \dots, t+\ell\) in that order, trying to build an entirely new red interval from scratch.
       - This generates \(\ell+2\) red intervals of length \(\ell+1\) in \(\ell+1\) turns.
     - If we repeat this procedure \(M\) times, we get a total production of \(M(\ell+2)\) red intervals of length \(\ell+1\) over \(M(\ell+1)\) turns.
     - Since Big Z can kill at most one red interval per turn, we have at least \(M\) red intervals left over at the end, hence at least \(M/2\) in either \(X\) or \(Y\).
     - Taking \(M\) large enough completes the induction.

6. **Conclusion:**
   - Suppose \(\{a, \dots, a+n-1\}\) is a red interval of length \(n\) in \(X\).
   - If \((a, y, z)\) is any point with a shoe on it, Little D can drop a shoe on each of \((a+1, y, z)\), \((a+2, y, z)\), \dots, \((a+n-1, y, z)\) and win.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) for all \(n\).