= Solution
For even $s$, suppose $K_{2s-1}$ had no monochromatic star with $s$ leaves. Both color degrees would be at most $s-1$ and their sum is $2s-2$, so both must equal $s-1$. The red graph would have odd degree at every one of an odd number of vertices, contradicting the <handshaking lemma>. Thus $2s-1$ vertices suffice.
On $2s-2$ vertices, split the vertices into two sets of size $s-1$, coloring inside each set red and across the sets blue. Red degrees are $s-2$ and blue degrees are $s-1$, so this avoids the star. Therefore
$$
\boxed{R(K_{1,s})=2s-1\quad(s\text{ even}).}
$$
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