1. Initial Setup and Pigeonhole Principle Application:
- We are given six points $P_1, P_2, P_3, P_4, P_5, P_6$ in 3-dimensional space such that no four of them lie in the same plane.
- Each line segment $P_jP_k$ is colored either black or white.
- We need to prove that there exists a triangle $P_jP_kP_l$ whose edges are all the same color.
- Consider one point, say $P_1$. There are five line segments from $P_1$ to the other points $P_2, P_3, P_4, P_5, P_6$.

2. Application of the Pigeonhole Principle:
- By the pigeonhole principle, among the five line segments $P_1P_2, P_1P_3, P_1P_4, P_1P_5, P_1P_6$, at least three of them must be of the same color. Without loss of generality, assume these three segments are $P_1P_2, P_1P_3, P_1P_4$ and they are colored black.

3. Formation of Tetrahedron and Analysis:
- Consider the tetrahedron formed by points $P_1, P_2, P_3, P_4$.
- We need to check the colors of the edges $P_2P_3, P_2P_4, P_3P_4$.

4. Case Analysis:
- If any of the edges $P_2P_3, P_2P_4, P_3P_4$ is black, then we have a monochromatic triangle. For example, if $P_2P_3$ is black, then $\triangle P_1P_2P_3$ is a monochromatic triangle.
- Suppose none of the edges $P_2P_3, P_2P_4, P_3P_4$ are black. This means all these edges are white.

5. Conclusion:
- If $P_2P_3, P_2P_4, P_3P_4$ are all white, then $\triangle P_2P_3P_4$ is a monochromatic triangle with all edges white.
- Therefore, in either case, we have found a monochromatic triangle.

$\blacksquare$