Consider the following 7 numbers: $2^1, 2^3, \ldots, 2^{13} = 8192 < 10000$. Clearly, the sum of any subset of them is such that the highest power of $2$ that divides the sum is odd, thus, it is not a perfect square.

Add number $3$ to the chosen numbers. Suppose it is possible to choose several numbers such that their sum is a perfect square. Then $3$ is one of such numbers, otherwise we get a contradiction as shown above. Since a perfect square can only have a remainder $0$, $1$ or $4$ when divided by $8$, and since the sum is odd, then the only possible remainder is $1$. However, only numbers $2$ and $3$ have a non-zero remainder when divided by $8$, thus, it is not possible to obtain a sum with remainder $1$ when divided by $8$.