Cockayne and Lorimer ("The Ramsey number for stripes" 1975) prove that in every $r$-colored complete graph on $n=\sum_{i=1}^rm_i+m_1-(r-1)$ vertices, where $m_1\geq \dots\geq m_r$, has a monochromatic matching in color $i$ of size $m_i$ (and this is tight).  When $m_1=\dots=m_r$, we get $n=(r+1)m-(r-1)$.  Their proof is a clever induction argument, but it seems to fundamentally use the fact that they are proving the more general ("non-diagonal") version.  There should be no complaints since they are able to prove the statement by proving something more general.  However, if you think about the problem for a few minutes, you will be able to come up with a proof of the 2-color case, and a few minutes more, you'll get the three color case, but I haven't found a proof for the $r$-color case which does not rely on proving the more general non-diagonal statement.  Basically, I'm curious if there is another proof of the diagonal version of their result along the lines of the $r=2$ and $r=3$ case below.  Beyond that, I'm wondering if there is a non-inductive proof which uses the Berge-Tutte theorem/formula (for instance) on the size of a largest matching in a graph.

Proof for $r=2$: We have $n=3m-1$, so if we can find a set of three vertices which contain both a red edge and a blue edge, we can apply induction on the remaining $n-3$ vertices.  If we can't find such a set, then only one color is used. 

Proof for $r=3$: We have $n=4m-2$, so if we can find a set of at most four vertices which contain edges of all three colors, we can apply induction on the remaining at least $n-4$ vertices.  If we can't find such a set, then we can deduce that only two colors are used.  Indeed, we are done if some vertex is incident with edges of all three colors or all vertices are incident with edges of only one color, so some vertex is incident with say only red and blue edges.  But now if there is a green edge anywhere, this will give a set of size at most 4 which sees edges of all three colors.