Problem 1749

Since $n$ is odd, $n^4 \equiv 1 \pmod{8}$. Since $x_i$ is odd, $x_i^2 \equiv 1 \pmod{8}$ for $1 \le i \le n$. Hence $n = x_1^2 + x_2^2 + \dots + x_n^2 \equiv n^4 \equiv 1 \pmod{8}$.

On the other hand, if $n \equiv 1 \pmod{8}$, 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
$$\begin{aligned} n^4 &= (8k + 1)^4 \\ &= (8k - 1)^4 + (8k + 1)^4 - (8k - 1)^4 \\ &= (8k - 1)^4 + ((8k + 1)^2 - (8k - 1)^2)((8k + 1)^2 + (8k - 1)^2) \\ &= (8k - 1)^4 + 32k(128k^2 + 2) \\ &= (8k - 1)^4 + 4k(32k - 1)^2 + (16k - 1)^2 + (92k - 1) \\ &= (8k - 1)^4 + 4k(32k - 1)^2 + (16k - 1)^2 + 92(k - 1) + 91 \\ &= ((8k - 1)^2)^2 + 4k(32k - 1)^2 + (16k - 1)^2 \\ &\qquad + (k - 1)(9^2 + 3^2 + 1^2 + 1^2) + (9^2 + 3^2 + 1^2) \end{aligned}$$
gives $n^4$ as a sum of $1 + 4k + 1 + 4(k-1) + 3 = 8k + 1 = n$ odd perfect squares.