Team Selection Test for IMO 2007

Since $n$ is odd, $n^4\equiv 1(\mathrm{mod}8)$. Since $x_i$ is odd, $x_i^2\equiv 1(\mathrm{mod}8)$ for $1\le i\le n$. Hence $n=x_1^2+x_2^2+\cdots +x_n^2\equiv n^4\equiv 1(\mathrm{mod}8)$.

On the other hand, if $n\equiv 1(\mathrm{mod}8)$, then odd numbers $x_1,x_2,\dots ,x_n$ satisfying the required equality can be found. If $n=1$, then $x_1=1$ will do: $n^4=1=x_1^2$. If $n=8k+1$ where $k$ is a positive integer, then $$\begin{array}{rl}{n}^4& =(8k+1{)}^4\\ & =(8k-1{)}^4+(8k+1{)}^4-(8k-1{)}^4\\ & =(8k-1{)}^4+((8k+1{)}^2-(8k-1{)}^2)((8k+1{)}^2+(8k-1{)}^2)\\ & =(8k-1{)}^4+32k(128k^2+2)\\ & =(8k-1{)}^4+4k(32k-1{)}^2+(16k-1{)}^2+(92k-1)\\ & =(8k-1{)}^4+4k(32k-1{)}^2+(16k-1{)}^2+92(k-1)+91\\ & =((8k-1{)}^2{)}^2+4k(32k-1{)}^2+(16k-1{)}^2\\ & +(k-1)(9^2+3^2+1^2+1^2)+(9^2+3^2+1^2)\end{array}$$ gives $n^4$ as a sum of $1+4k+1+4(k-1)+3=8k+1=n$ odd perfect squares.