No. of masters = No. of juniors = $2*15$. Now $15$ is odd. We will prove the assertion for $(2n-1)$ rather than $15$, using Graph Theoretic Language. Let no. of masters and juniors be $2(2n-1)$. Consider two disjoint graphs $G$ and $G'$ each having $2(2n-1)$ vertices. $G$ represents the masters and $G'$ represents the juniors. Now each master plays with exactly one master so each vertex has degree exactly $1$. This implies that $G$ breaks up into $(2n-1)$ disjoint pairs of connected vertices. Similarly $G'$ also breaks into $(2n-1)$ disjoint pairs of connected vertices. Any vertex $A$ in $G$ is exactly connected to $(2n-1)$ vertices of $G'$ as each master plays with $(2n-1)$ juniors. Call these vertices in $G'$ as the splinter of the vertex $A$. Now the problem statement transforms to proving that the union of splinters of $2$ connected vertices in $G$ contains a pair of connected vertices in $G'$. Suppose this is not true. Then the union of splinters consists of at most $(2n-1)$ vertices, one vertex from each connected pair of $G'$. Since the splinter of a vertex in $G$ has cardinal no. $(2n-1)$ so the union of two splinters has cardinal no. $\geq(2n-1)$ with equality iff the two splinters are exactly the same. Now equality is reached for each pair of connected vertices in $G$ since their union of splinters has maximum cardinal no. $(2n-1)$, by earlier observation. So each connected pair in $G$ has the same splinter in $G'$. Consider a vertex in $G'$. It is either connected to both the vertices of a connected pair in $G$ or to none of them otherwise the equality of their splinters is violated. This implies that the vertex in $G'$ is connected to an integer no. of connected pairs in $G$, that is, to an even no. of vertices in $G$ as each connected pair has two vertices. But this contradicts the fact that a vertex in $G'$ is connected to exactly $(2n-1)$ vertices, that is, odd no. of vertices in $G$. This contradiction ends the proof thereby giving the desired result.

In your argument, you need to veriry <babu2001> wrote: Now the problem statement transforms to proving that the union of splinters of $2$ connected vertices in $G$ contains a pair of connected vertices in $G'$. To see this, color each vertex of each pair of connected masters and juniors with black and white marker. Even though the union of the vertices of a connected component in $G$ contains a connected component in $G'$, say $g$, it is still possible that only splinter of white vertex contains $g$ where the splinter of the black vertex is disjoint with $g$.

Let the pair of two junior players matched each other be $J_{1},J_{2},...,J_{15}$ as well as the pair of two masters be $M_{1},M_{2},...,M_{15}$. We will make use of reductio ad absurdum. Suppose that every $J_{i},M_{i}$ be vertices of graph, draw edge between $J_{i}$ and $M_{k}$ if there is a match. If there are 3 edges or more between $J_{i}$ and $M_{k}$, contradiction. so every pair of $(J_{i},M_{k})$ have 2 edges considering total master-junior games equal to $450=15\times15\times2$. By this condition, for every pair of $(J_{i},M_{k})$, every edges connected to single junior $j\subset J_{i}$ or single master $m\subset M_{k}$. Let the number of former pairs be $x$, later be $y$ which satisfies $x+y=225$. with an arbitrary $J_{i}$, the number of $M_{k}$ which connected 2 edges in $m\subset M_{k}$ is odd because one junior player matched masters 15 times. So $y$ should be odd. this condition holds in $x$ at that time. Contradiction.