Let these $21$ entrances be $A_1,A_2,...,A_{21}$ in a clockwise order. Since no three walkways meet at a single point, there are $9\cdot10=90$ plants on walkway $A_1A_{11}$. By pigeonhole principle., there are at least $\left\lceil \frac{90}{2} \right\rceil=45$ same type of plant on $A_1A_{11}$. Thus walkway $A_1A_{11}$ is decorated with lights. Similarly, walkway $A_2A_{12},A_3A_{13},...,A_{21}A_{10}$ are also decorated with lights. So, there are at least $21$ walkways that are decorated with lights. By pigeonhole principle., there are at least $\left\lceil \frac{21}{2} \right\rceil=11$ walkways that are decorated with lights and paved with the same material.

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Label the $21$ entrances by $1,2,\dots,21$ in that order and a walkway joining entrances $a,b$ by $(a,b)$. Note that the walkways$(n,n+10)$ and $(n,n+11)$ has exactly $9\times 10=90$ plants. By the Pigeonhole principle, There exist one type of plant with at least $45$ plants in that walkway. So that walkway is decorated with lights. Lastly, There are at least $21$ walkways $(1,11),(2,12),\dots,(21,10)$ decorated with lights. Again, by the pigeonhole principle There are at least $11$ walkways decorated with lights and paved with the same material.

Let the entrances be $A_0,A_1,\dots,A_{20}.$ Consider the pathway $A_iA_{i+10},$ where indices are taken modulo $21.$ There are $9\cdot 10$ other paths that intersect with $A_iA_{i+10}$ so at least $45$ of the paths must be the same or at least $45$ of the paths must be different by pigeonhole; in either case, at least $45$ of the same flower is planted on the path so the path is lit. Notice this is true for all $A_iA_{i+10}$ for $0\le i\le 10$ and for all $A_iA_{i+11}$ for $0\le i\le 9.$ Hence, there are $21$ paths with lights, and by pigeonhole at least $11$ must be of the same material. $\square$