Find the largest possible value of $a$ such that there exist real numbers $b,c>1$ such that
        \[a^{\log_b c}\cdot b^{\log_c a}=2023.\]

[i]Proposed by Howard Halim[/i]
1. **Substitute expressions for \( b \) and \( c \):**
   Let \( c = b^y \) and \( b = a^x \). Then, the given equation becomes:
   \[
   a^{\log_b c} \cdot b^{\log_c a} = 2023
   \]
   Since \( \log_b c = y \) and \( \log_c a = \frac{1}{\log_a c} = \frac{1}{xy} \), the equation transforms to:
   \[
   a^y \cdot b^{\frac{1}{xy}} = 2023
   \]

2. **Express \( b \) in terms of \( a \):**
   Given \( b = a^x \), substitute this into the equation:
   \[
   a^y \cdot (a^x)^{\frac{1}{xy}} = 2023
   \]
   Simplify the exponent of \( a \):
   \[
   a^y \cdot a^{\frac{1}{y}} = 2023
   \]
   Combine the exponents:
   \[
   a^{y + \frac{1}{y}} = 2023
   \]

3. **Apply the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know:
   \[
   y + \frac{1}{y} \geq 2
   \]
   This implies:
   \[
   a^{y + \frac{1}{y}} \geq a^2
   \]
   Since \( a^{y + \frac{1}{y}} = 2023 \), we have:
   \[
   a^2 \leq 2023
   \]

4. **Solve for \( a \):**
   Taking the square root of both sides, we get:
   \[
   a \leq \sqrt{2023}
   \]

5. **Determine the largest possible value of \( a \):**
   The largest possible value of \( a \) that satisfies the inequality is:
   \[
   a = \sqrt{2023}
   \]

The final answer is \(\boxed{\sqrt{2023}}\).