If $N$ is the number of triangles of different shapes (i.e., not similar) whose angles are all integers (in degrees), what is $\frac{N}{100}$?
1. **Understanding the Problem:**
   We need to find the number of triangles with integer angles that are not similar to each other. The sum of the angles in any triangle is \(180^\circ\). We need to count the number of distinct sets of angles \((a, b, c)\) where \(a, b, c\) are integers, \(a \leq b \leq c\), and \(a + b + c = 180\).

2. **Fixing One Angle:**
   Let's fix \(a\) and find the number of possible pairs \((b, c)\) such that \(a + b + c = 180\) and \(b \leq c\). This means \(b + c = 180 - a\).

3. **Counting Solutions:**
   For each fixed \(a\), the number of solutions \((b, c)\) where \(b \leq c\) is given by:
   \[
   \left\lfloor \frac{180 - a}{2} \right\rfloor
   \]
   This is because \(b\) can range from \(a\) to \(\left\lfloor \frac{180 - a}{2} \right\rfloor\).

4. **Summing Over All Possible Values of \(a\):**
   We need to sum the number of solutions for \(a\) from 1 to 59 (since \(a\) must be less than 60 to ensure \(b \leq c\)):
   \[
   \sum_{a=1}^{59} \left\lfloor \frac{180 - a}{2} \right\rfloor
   \]

5. **Simplifying the Sum:**
   Notice that:
   \[
   \left\lfloor \frac{180 - a}{2} \right\rfloor = \left\lfloor 90 - \frac{a}{2} \right\rfloor
   \]
   We can split this into two cases based on whether \(a\) is even or odd.

6. **Calculating the Sum:**
   For even \(a = 2k\):
   \[
   \left\lfloor 90 - k \right\rfloor = 90 - k
   \]
   For odd \(a = 2k + 1\):
   \[
   \left\lfloor 90 - \frac{2k + 1}{2} \right\rfloor = \left\lfloor 90 - k - 0.5 \right\rfloor = 89 - k
   \]

   Summing these:
   \[
   \sum_{k=0}^{29} (90 - k) + \sum_{k=0}^{29} (89 - k)
   \]

7. **Evaluating the Sums:**
   \[
   \sum_{k=0}^{29} (90 - k) = 90 \times 30 - \sum_{k=0}^{29} k = 2700 - \frac{29 \times 30}{2} = 2700 - 435 = 2265
   \]
   \[
   \sum_{k=0}^{29} (89 - k) = 89 \times 30 - \sum_{k=0}^{29} k = 2670 - 435 = 2235
   \]

8. **Total Number of Triangles:**
   \[
   N = 2265 + 2235 = 4500
   \]

9. **Finding \(\frac{N}{100}\):**
   \[
   \frac{N}{100} = \frac{4500}{100} = 45
   \]

The final answer is \(\boxed{45}\).