Let $P$ be a $2019-$gon, such that no three of its diagonals concur at an internal point. We will call each internal intersection point of diagonals of $P$ a knot. What is the greatest number of knots one can choose, such that there doesn't exist a cycle of chosen knots? ( Every two adjacent knots in a cycle must be on the same diagonal and on every diagonal there are at most two knots from a cycle.)
1. **Understanding the Problem:**
   We need to find the maximum number of knots (intersection points of diagonals) in a 2019-gon such that no cycle is formed. A cycle is formed if there is a closed loop of knots where each pair of adjacent knots in the cycle lies on the same diagonal.

2. **Counting the Total Number of Knots:**
   The number of diagonals in an \( n \)-gon is given by:
   \[
   \frac{n(n-3)}{2}
   \]
   For \( n = 2019 \):
   \[
   \text{Number of diagonals} = \frac{2019 \times 2016}{2} = 2019 \times 1008
   \]
   Each pair of diagonals intersects at most once inside the polygon. The number of intersection points (knots) can be calculated using the combination formula \( \binom{n}{4} \):
   \[
   \binom{2019}{4} = \frac{2019 \times 2018 \times 2017 \times 2016}{4 \times 3 \times 2 \times 1}
   \]

3. **Avoiding Cycles:**
   To avoid cycles, we need to ensure that no set of knots forms a closed loop. This can be visualized as a graph problem where each knot is a vertex and each diagonal segment between knots is an edge. We need to find the maximum number of vertices (knots) such that the graph is acyclic (a forest).

4. **Graph Theory Insight:**
   In graph theory, a forest (a collection of trees) with \( V \) vertices has at most \( V-1 \) edges. To maximize the number of vertices while ensuring no cycles, we need to maximize the number of vertices in a forest.

5. **Maximizing the Number of Knots:**
   Since each diagonal can intersect with many others, we need to ensure that no three diagonals intersect at the same point. This means each intersection point (knot) is unique to a pair of diagonals.

6. **Conclusion:**
   The maximum number of knots that can be chosen without forming a cycle is the number of vertices in a maximal forest. For a 2019-gon, the number of such knots is:
   \[
   \boxed{2018}
   \]