How many solutions does the equation:
$$[\frac{x}{20}]=[\frac{x}{17}]$$
have over the set of positve integers?
$[a]$ denotes the largest integer that is less than or equal to $a$.
[i]Proposed by Karl Czakler[/i]
To solve the equation \(\left\lfloor \frac{x}{20} \right\rfloor = \left\lfloor \frac{x}{17} \right\rfloor\) over the set of positive integers, we need to find the values of \(x\) for which the floor functions of \(\frac{x}{20}\) and \(\frac{x}{17}\) are equal.

1. Let \(k = \left\lfloor \frac{x}{20} \right\rfloor = \left\lfloor \frac{x}{17} \right\rfloor\). This implies:
   \[
   k \leq \frac{x}{20} < k+1 \quad \text{and} \quad k \leq \frac{x}{17} < k+1
   \]

2. From the inequalities, we can derive:
   \[
   20k \leq x < 20(k+1) \quad \text{and} \quad 17k \leq x < 17(k+1)
   \]

3. To find the intersection of these ranges, we take the maximum of the lower bounds and the minimum of the upper bounds:
   \[
   \max(20k, 17k) \leq x < \min(20(k+1), 17(k+1))
   \]
   Since \(20k \geq 17k\), the lower bound is \(20k\). Similarly, since \(20(k+1) > 17(k+1)\), the upper bound is \(17(k+1)\).

4. Therefore, the range for \(x\) is:
   \[
   20k \leq x < 17(k+1)
   \]

5. To find the number of integers \(x\) in this range, we calculate the length of the interval:
   \[
   17(k+1) - 20k = 17k + 17 - 20k = 17 - 3k
   \]

6. We need to sum these lengths for all possible values of \(k\) such that the interval is non-empty. The interval is non-empty as long as \(17 - 3k > 0\), which gives:
   \[
   17 > 3k \implies k < \frac{17}{3} \approx 5.67
   \]
   Thus, \(k\) can take integer values from 0 to 5.

7. Summing the lengths for \(k = 0\) to \(k = 5\):
   \[
   \sum_{k=0}^{5} (17 - 3k)
   \]

8. Calculating the sum:
   \[
   (17 - 3 \cdot 0) + (17 - 3 \cdot 1) + (17 - 3 \cdot 2) + (17 - 3 \cdot 3) + (17 - 3 \cdot 4) + (17 - 3 \cdot 5)
   \]
   \[
   = 17 + 14 + 11 + 8 + 5 + 2
   \]
   \[
   = 57
   \]

9. Since we are considering positive integers, we exclude \(x = 0\) if it falls within the range. However, in this case, \(x = 0\) is not a positive integer, so we do not need to subtract 1.

The final answer is \(\boxed{57}\).