Denote the card with $i, j$ written on it as $\{i, j\}$. It is easy to know that these 10 cards are exactly $\{i, j\}$ ($1 \le i < j \le 5$).

Consider the “good” placements of the cards. There are 10 cards in the five boxes, so there are at least 3 cards in box 1. The only cards that will fit in box 1 are $\{1, 2\}$, $\{1, 3\}$, $\{1, 4\}$ and $\{1, 5\}$.

Case 1: The 4 cards are all placed in box 1, and at this point it is no longer possible to have 4 cards in each of the remaining boxes. Therefore, no matter how the remaining 6 cards are placed, they will fit the requirements and there are $2^6 = 64$ “good” placements.

Case 2: There are exactly 3 of the 4 cards in box 1 and the rest of each box contains at most 2 cards.

Consider the number $N$ of placements of $\{1, 2\}$, $\{1, 3\}$, $\{1, 4\}$ in box 1 and $\{1, 5\}$ in box 5.

There are 8 possible ways to place the cards $\{2, 3\}$, $\{2, 4\}$, $\{3, 4\}$, with 6 of them being two cards placed in one of the boxes 2, 3, 4, and the remaining 2 being one card placed in each box of 2, 3, 4.

If there are two cards of $\{2, 3\}$, $\{2, 4\}$ and $\{3, 4\}$ in a box, suppose that $\{2, 3\}$ and $\{2, 4\}$ are in box 2, and then $\{2, 5\}$ can only be in box 5. Therefore, box 5 already has $\{1, 5\}$ and $\{2, 5\}$, so $\{3, 5\}$ and $\{4, 5\}$ are in boxes 3 and 4 respectively, namely, the placement of $\{2, 5\}$, $\{3, 5\}$ and $\{4, 5\}$ is unique.

If one card of $\{2, 3\}$, $\{2, 4\}$ and $\{3, 4\}$ is placed in each box of 2, 3, 4, then there are at most 2 cards in each box of 2, 3, 4. It is only necessary to make sure that there are no more than 2 cards in box 5, namely, there are 0 or 1 card of $\{2, 5\}$, $\{3, 5\}$ and $\{4, 5\}$ in box 5, and the number of the corresponding placements is $C_3^0 + C_3^1 = 4$.

As a result, $N = 6 \times 1 + 2 \times 4 = 14$. By symmetry, there are $4N = 56$ “good” placements in case 2.

To sum up, there is a total of $64 + 56 = 120$ “good” placements.