Find the largest real number $a$ such that \[\left\{ \begin{array}{l}
x - 4y = 1 \\
ax + 3y = 1\\
 \end{array} \right.
\] has an integer solution.
1. We start with the given system of linear equations:
   \[
   \begin{cases}
   x - 4y = 1 \\
   ax + 3y = 1
   \end{cases}
   \]

2. To eliminate \( y \), we can manipulate the equations. Multiply the first equation by 3 and the second equation by 4:
   \[
   3(x - 4y) = 3 \implies 3x - 12y = 3
   \]
   \[
   4(ax + 3y) = 4 \implies 4ax + 12y = 4
   \]

3. Add the two resulting equations:
   \[
   3x - 12y + 4ax + 12y = 3 + 4
   \]
   Simplifying, we get:
   \[
   3x + 4ax = 7
   \]
   \[
   x(3 + 4a) = 7
   \]

4. For \( x \) to be an integer, \( 3 + 4a \) must be a divisor of 7. The divisors of 7 are \( \pm 1 \) and \( \pm 7 \).

5. Consider the possible values for \( 3 + 4a \):
   - If \( 3 + 4a = 1 \):
     \[
     4a = 1 - 3 \implies 4a = -2 \implies a = -\frac{1}{2}
     \]
   - If \( 3 + 4a = -1 \):
     \[
     4a = -1 - 3 \implies 4a = -4 \implies a = -1
     \]
   - If \( 3 + 4a = 7 \):
     \[
     4a = 7 - 3 \implies 4a = 4 \implies a = 1
     \]
   - If \( 3 + 4a = -7 \):
     \[
     4a = -7 - 3 \implies 4a = -10 \implies a = -\frac{5}{2}
     \]

6. Among these values, the largest real number \( a \) is \( 1 \).

The final answer is \(\boxed{1}\).