Find all real numbers $a$ such that $3 < a < 4$ and $a(a-3\{a\})$ is an integer.

(Here $\{a\}$ denotes the fractional part of $a$.)
1. **Understanding the problem:**
   We need to find all real numbers \( a \) such that \( 3 < a < 4 \) and \( a(a - 3\{a\}) \) is an integer. Here, \(\{a\}\) denotes the fractional part of \( a \).

2. **Expressing \( a \) in terms of its integer and fractional parts:**
   Since \( 3 < a < 4 \), we can write \( a = 3 + x \) where \( 0 < x < 1 \). Here, \( x \) is the fractional part of \( a \), i.e., \( \{a\} = x \).

3. **Substituting \( a \) in the given expression:**
   We need to evaluate \( a(a - 3\{a\}) \):
   \[
   a(a - 3\{a\}) = (3 + x)\left((3 + x) - 3x\right) = (3 + x)(3 - 2x)
   \]

4. **Simplifying the expression:**
   \[
   (3 + x)(3 - 2x) = 3(3 - 2x) + x(3 - 2x) = 9 - 6x + 3x - 2x^2 = 9 - 3x - 2x^2
   \]
   We need \( 9 - 3x - 2x^2 \) to be an integer.

5. **Setting up the equation:**
   Let \( 9 - 3x - 2x^2 = k \) where \( k \) is an integer. Rearranging, we get:
   \[
   2x^2 + 3x = 9 - k
   \]
   Since \( x \in (0, 1) \), \( 2x^2 + 3x \) must be in the interval \( (0, 5) \). Therefore, \( 9 - k \) must be in the interval \( (0, 5) \), which implies \( k \) must be in the interval \( (4, 9) \). Since \( k \) is an integer, \( k \) can be \( 5, 6, 7, \) or \( 8 \).

6. **Solving for \( x \) for each \( k \):**
   - For \( k = 8 \):
     \[
     2x^2 + 3x = 1 \implies 2x^2 + 3x - 1 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}
     \]
     Since \( 0 < x < 1 \), we take the positive root:
     \[
     x = \frac{\sqrt{17} - 3}{4}
     \]

   - For \( k = 7 \):
     \[
     2x^2 + 3x = 2 \implies 2x^2 + 3x - 2 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}
     \]
     Since \( 0 < x < 1 \), we take the positive root:
     \[
     x = \frac{1}{2}
     \]

   - For \( k = 6 \):
     \[
     2x^2 + 3x = 3 \implies 2x^2 + 3x - 3 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 24}}{4} = \frac{-3 \pm \sqrt{33}}{4}
     \]
     Since \( 0 < x < 1 \), we take the positive root:
     \[
     x = \frac{\sqrt{33} - 3}{4}
     \]

   - For \( k = 5 \):
     \[
     2x^2 + 3x = 4 \implies 2x^2 + 3x - 4 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}
     \]
     Since \( 0 < x < 1 \), we take the positive root:
     \[
     x = \frac{\sqrt{41} - 3}{4}
     \]

7. **Finding the corresponding \( a \) values:**
   - For \( x = \frac{\sqrt{17} - 3}{4} \), \( a = 3 + \frac{\sqrt{17} - 3}{4} \)
   - For \( x = \frac{1}{2} \), \( a = 3 + \frac{1}{2} = \frac{7}{2} \)
   - For \( x = \frac{\sqrt{33} - 3}{4} \), \( a = 3 + \frac{\sqrt{33} - 3}{4} \)
   - For \( x = \frac{\sqrt{41} - 3}{4} \), \( a = 3 + \frac{\sqrt{41} - 3}{4} \)

The final answer is \( \boxed{ a = 3 + \frac{\sqrt{17} - 3}{4}, \frac{7}{2}, 3 + \frac{\sqrt{33} - 3}{4}, 3 + \frac{\sqrt{41} - 3}{4} } \).