Evan has $10$ cards numbered $1$ through $10$. He chooses some of the cards and takes the product of the numbers on them. When the product is divided by $3$, the remainder is $1$. Find the maximum number of cards he could have chose.

[i]Proposed by Evan Chang [/i]
1. **Identify the cards that cannot be chosen:**
   - The cards are numbered from $1$ to $10$.
   - We need the product of the chosen cards to leave a remainder of $1$ when divided by $3$.
   - Any card that is a multiple of $3$ will make the product divisible by $3$. Therefore, we cannot choose cards numbered $3$, $6$, or $9$.

2. **List the remaining cards:**
   - The remaining cards are $1, 2, 4, 5, 7, 8, 10$.

3. **Calculate the product modulo $3$:**
   - We need to find the product of these cards modulo $3$.
   - First, we find the modulo $3$ of each card:
     \[
     1 \equiv 1 \pmod{3}, \quad 2 \equiv 2 \pmod{3}, \quad 4 \equiv 1 \pmod{3}, \quad 5 \equiv 2 \pmod{3}, \quad 7 \equiv 1 \pmod{3}, \quad 8 \equiv 2 \pmod{3}, \quad 10 \equiv 1 \pmod{3}
     \]

4. **Calculate the product of these residues modulo $3$:**
   - The product of the residues is:
     \[
     1 \cdot 2 \cdot 1 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 2 \cdot 2 \cdot 2 = 8
     \]
   - Now, find $8 \pmod{3}$:
     \[
     8 \div 3 = 2 \text{ remainder } 2 \quad \Rightarrow \quad 8 \equiv 2 \pmod{3}
     \]

5. **Adjust the selection to achieve the desired remainder:**
   - We need the product to be $1 \pmod{3}$, but the current product is $2 \pmod{3}$.
   - To change the product from $2 \pmod{3}$ to $1 \pmod{3}$, we can remove one card that contributes $2 \pmod{3}$.
   - The cards contributing $2 \pmod{3}$ are $2, 5, 8$.

6. **Remove one card contributing $2 \pmod{3}$:**
   - Removing any one of $2, 5, 8$ will change the product to $1 \pmod{3}$.
   - For example, if we remove the card numbered $2$, the remaining cards are $1, 4, 5, 7, 8, 10$.
   - The product of these cards modulo $3$ is:
     \[
     1 \cdot 1 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 2 \cdot 2 = 4 \equiv 1 \pmod{3}
     \]

7. **Verify the maximum number of cards:**
   - By removing one card from the set of $7$ cards, we are left with $6$ cards.
   - Therefore, the maximum number of cards that can be chosen such that their product leaves a remainder of $1$ when divided by $3$ is $6$.

The final answer is $\boxed{6}$.