Let $v_1<v_2<\dots <v_{10}$ be the velocities of the participants and $L$ be the length of the track. Any two participants $i, j$ with $i<j$ meet for the first time at time $t=\frac{L}{v_j-v_i}$. Let $r=\min{(v_{k+1}-v_k)}$ with $1\le k\le9$: the participants for which this minimum is attained will be the last to meet. Hence, any two participants $i, j$ will meet $\left\lfloor{\frac{v_j-v_i}{r}}\right\rfloor$ times. We wish to prove that for a fixed $a$ we have $\sum_{i=1}^{10}{\left\lfloor{\frac{|v_a-v_i|}{r}}\right\rfloor\ge 25}$. It's easy to see that $|v_j-v_i|\ge |j-i|r$. So $\sum_{i=1}^{10}{\left\lfloor{\frac{|v_a-v_i|}{r}}\right\rfloor\ge\sum_{i=1}^{10}|a-i|}$. For $a=5$ we obtain $25$, and for $a\neq5$ we obtain more than $25$.

My proof (which is very similar to Motal) also works for $2n$ cyclists and $n^2 $ meeting .