A chocolate bar in the shape of an equilateral triangle with side of the length $n$, consists of triangular chips with sides of the length $1$, parallel to sides of the bar. Two players take turns eating up the chocolate. Each player breaks off a triangular piece (along one of the lines), eats it up and passes leftovers to the other player (as long as bar contains more than one chip, the player is not allowed to eat it completely).

A player who has no move or leaves exactly one chip to the opponent, loses. For each $n$, find who has a winning strategy.
1. **Initial Setup**: Consider a chocolate bar in the shape of an equilateral triangle with side length \( n \). The bar consists of smaller triangular chips with side length \( 1 \), arranged such that they are parallel to the sides of the bar.

2. **Game Rules**: Two players take turns breaking off a triangular piece along one of the lines and eating it. The player who cannot make a move or leaves exactly one chip for the opponent loses.

3. **Case \( n = 1 \)**: If \( n = 1 \), the chocolate bar consists of a single triangular chip. The first player has no move and loses immediately.

4. **Case \( n > 1 \)**: For \( n > 1 \), the first player can adopt the following strategy:
   - Break off a triangle of side \( n-1 \). This leaves a trapezoidal piece made of \( 2n-1 \) smaller triangles of side \( 1 \).

5. **Remaining Moves**: After the first move, the remaining piece is a trapezoid with \( 2n-1 \) triangles. Each player can only break off a triangle of side \( 1 \) in subsequent moves.

6. **Counting Moves**: 
   - The total number of smaller triangles in the original equilateral triangle is given by the formula for the sum of the first \( n \) triangular numbers:
     \[
     T_n = \frac{n(n+1)}{2}
     \]
   - After the first move, the number of remaining triangles is \( 2n-1 \).

7. **Game Progression**: 
   - Each player takes turns breaking off one triangle of side \( 1 \). 
   - After \( 2n-2 \) moves, there will be exactly one triangle left.

8. **Winning Strategy**: 
   - The first player makes the initial move, leaving \( 2n-1 \) triangles.
   - After \( 2n-2 \) moves, the second player will face the last remaining triangle and will lose because they cannot make a move.

Conclusion:
The first player has a winning strategy for all \( n > 1 \) by breaking off a triangle of side \( n-1 \) on the first move and then continuing to break off triangles of side \( 1 \) in subsequent moves.

The final answer is the first player has a winning strategy for all \( \boxed{ n > 1 } \).