, 2012

Answer: any integer number from $1$ to $[N/2]$.

Consider arbitrary arrangement of the boys along the circle. We put the signs "+" or "-" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign "+" before the boy if he is taller than the previous boy and we put the sign "-" if he is shorter than the previous one. It is evident that the boy is tall if and only if the sign "+" stands before him and the sign "-" stands after him. (Note that since the tallest boy among all $N$ boys is tall, there exist the signs "+" as well as the signs "-" in any arrangement.) Therefore, the number of tall boys in the arrangement is equal to the number of alternations of "+" and "-". It is evident that the number of these alternations is less than or equal to $[N/2]$.

Now we show that for any $b$, $1 \le b \le [N/2]$, there exists an arrangement with exactly $b$ tall boys.

We number all boys in accordance with their heights: the shortest boy has the number $1$, and the tallest boy has the number $N$. We partition all boys into three groups $A, B$, and $C$: group $A$ contains the shortest boys, i.e. the boys with the numbers $1, 2, \dots, b$; group $B$ contains the tallest boys, i.e. the boys with the numbers $N-b+1, N-b+2, \dots, N$ (since $b \le [N/2]$, i.e. $2b \le N$, there exists such partition); finally, group $C$ consists of all remaining boys (if $N$ is even and $b = N/2$, then $C$ is empty).

We also number the places on the circle with numbers from $1$ to $N$. We place the boys from $A$ on the places with the numbers $2k-1$, $k=1, \dots, b$; the boys from $B$ are placed on the places with the numbers $2k$, $k=1, \dots, b$; the boys from $C$ are placed on the remaining places (with the numbers $2b+1, \dots, N$) so that the boy with the number $N-b$ is placed on the place with the number $2b+1, \dots$, the boy with the number $b+1$ is placed on the place with the number $N$ (see the fig.). It is easy to see that there exist exactly $b$ tall boys in this arrangement.