There are $ 7$ boxes arranged in a row and numbered $1$ through $7$. You have a stack of $2015$ cards, which you place one by one in the boxes. The first card is placed in box #$1$, the second in box #$2$, and so forth up to the seventh card which is placed in box #$7$.  You then start working back in the other direction, placing the eighth card in box #$6$, the ninth in box #$5$, up to the thirteenth card being placed in box #$1$. The fourteenth card is then placed in box #$2$, and this continues until every card is distributed. What box will the last card be placed in?
1. We start by understanding the pattern of card placement. The cards are placed in a zigzag manner:
   - Cards 1 to 7 are placed in boxes 1 to 7.
   - Cards 8 to 13 are placed in boxes 6 to 1.
   - Cards 14 to 20 are placed in boxes 2 to 7.
   - Cards 21 to 27 are placed in boxes 6 to 1.
   - This pattern repeats every 12 cards.

2. To determine the box for the 2015th card, we need to find the position of 2015 within the repeating cycle of 12 cards. We do this by calculating the remainder when 2015 is divided by 12:
   \[
   2015 \div 12 = 167 \text{ remainder } 11
   \]
   This means that 2015 is 11 cards into the 168th cycle.

3. We now need to determine the box for the 11th card in the cycle:
   - The first 7 cards in the cycle go into boxes 1 to 7.
   - The next 5 cards (8 to 12) go into boxes 6 to 2.

4. Since 11 is within the second part of the cycle (8 to 12), we count backwards from box 6:
   - 8th card goes to box 6
   - 9th card goes to box 5
   - 10th card goes to box 4
   - 11th card goes to box 3

Therefore, the 2015th card will be placed in box 3.

The final answer is $\boxed{\text{box 3}}$