Problem 1474

To solve this problem, we need to ensure that for all $1 \le i < j \le 10$, if $i$ divides $j$, then the $i$-th and $j$-th squares have different colors. We will use a step-by-step approach to count the number of valid colorings.

1. Color the 1st square:
- There are 4 possible colors for the 1st square.

2. Color the 7th square:
- The 7th square is not influenced by any other square because 7 is a prime number and does not divide any other number in the set $\{1, 2, \ldots, 10\}$. Thus, there are 3 remaining colors for the 7th square.

3. Color the remaining squares:
- We need to consider the squares that are influenced by the divisibility condition. These squares are $2, 3, 4, 5, 6, 8, 9, 10$.

4. **Casework on $2, 3, 5$:**
- We will consider different cases based on the colors of squares $2, 3, 5$.

**Case 1: $2 \neq 3$ and $2 \neq 5$:**
- There are 3 choices for the color of square 2.
- There are 2 choices for the color of square 3 (since it must be different from square 2).
- There are 2 choices for the color of square 5 (since it must be different from square 2).
- Thus, there are $3 \times 2 \times 2 = 12$ ways to color squares $2, 3, 5$.

**Case 2: $2 = 3 \neq 5$:**
- There are 3 choices for the color of square 2 (which also determines the color of square 3).
- There are 2 choices for the color of square 5.
- Thus, there are $3 \times 2 = 6$ ways to color squares $2, 3, 5$.

**Case 3: $2 = 5 \neq 3$:**
- There are 3 choices for the color of square 2 (which also determines the color of square 5).
- There are 2 choices for the color of square 3.
- Thus, there are $3 \times 2 = 6$ ways to color squares $2, 3, 5$.

**Case 4: $2 = 3 = 5$:**
- There are 3 choices for the color of squares $2, 3, 5$.
- Thus, there are $3$ ways to color squares $2, 3, 5$.

5. **Color the remaining squares $4, 6, 8, 9, 10$:
- For each of the above cases, we need to ensure that the remaining squares are colored such that the divisibility condition is satisfied.
- For each case, we have:
- 1 choice for square 4 (since it must be different from square 2).
- 1 choice for square 6 (since it must be different from squares 2 and 3).
- 1 choice for square 8 (since it must be different from square 4).
- 1 choice for square 9 (since it must be different from square 3).
- 1 choice for square 10 (since it must be different from squares 2 and 5).

6. Total number of ways:**
- Summing up the number of ways for each case:
- Case 1: $12$ ways
- Case 2: $6$ ways
- Case 3: $6$ ways
- Case 4: $3$ ways
- Total: $12 + 6 + 6 + 3 = 27$ ways

7. Multiplying with the factors we ignored:
- We initially ignored the factors for coloring the 1st and 7th squares.
- Thus, the total number of ways is $27 \times 4 \times 3 = 324$.

The final answer is $\boxed{324}$.