Let $x, y$ and $z$ be consecutive integers such that
\[\frac 1x+\frac 1y+\frac 1z >\frac{1}{45}.\]
Find the maximum value of $x + y + z$.
1. Let \( x, y, z \) be consecutive integers such that \( x < y < z \). We can write \( x = y-1 \) and \( z = y+1 \).
2. The given inequality is:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} > \frac{1}{45}
   \]
   Substituting \( x = y-1 \) and \( z = y+1 \), we get:
   \[
   \frac{1}{y-1} + \frac{1}{y} + \frac{1}{y+1} > \frac{1}{45}
   \]
3. To find the maximum value of \( x + y + z \), we need to find the largest \( y \) such that the inequality holds. Let's test \( y = 135 \):
   \[
   \frac{1}{134} + \frac{1}{135} + \frac{1}{136}
   \]
4. We need to check if:
   \[
   \frac{1}{134} + \frac{1}{135} + \frac{1}{136} > \frac{1}{45}
   \]
5. To compare these fractions, we can use the approximation that for large \( n \), \( \frac{1}{n} \approx \frac{1}{n+1} \approx \frac{1}{n-1} \). Thus:
   \[
   \frac{1}{134} + \frac{1}{135} + \frac{1}{136} \approx \frac{3}{135}
   \]
6. We need to check if:
   \[
   \frac{3}{135} > \frac{1}{45}
   \]
   Simplifying, we get:
   \[
   \frac{3}{135} = \frac{1}{45}
   \]
   This shows that \( y = 135 \) is the boundary case. To ensure the inequality is strict, we need to check slightly smaller values of \( y \).

7. Let's test \( y = 134 \):
   \[
   \frac{1}{133} + \frac{1}{134} + \frac{1}{135}
   \]
8. We need to check if:
   \[
   \frac{1}{133} + \frac{1}{134} + \frac{1}{135} > \frac{1}{45}
   \]
9. Using the same approximation:
   \[
   \frac{1}{133} + \frac{1}{134} + \frac{1}{135} \approx \frac{3}{134}
   \]
10. We need to check if:
    \[
    \frac{3}{134} > \frac{1}{45}
    \]
    Simplifying, we get:
    \[
    \frac{3}{134} \approx 0.0224 \quad \text{and} \quad \frac{1}{45} \approx 0.0222
    \]
    Since \( 0.0224 > 0.0222 \), the inequality holds for \( y = 134 \).

11. Therefore, the maximum value of \( x + y + z \) is when \( y = 134 \):
    \[
    x + y + z = (y-1) + y + (y+1) = 3y = 3 \cdot 134 = 402
    \]

The final answer is \(\boxed{402}\)