By taking modulo $4$ we easily obtain that $n$ is either $0$ or $1$ mod 4. For $n=4 \cdot k +1$ with $k$ a nonnegative integer take : $ a_1 = n $ and let $2k$ of the other numbers be $-1$ and the rest ( $2k$ ) be $1$. For $n=8 \cdot k$ with $k$ a positive integer take : $ a_1 = 2 , a_2 = 4k $ and let $6k-2$ of the other numbers be $1$ and the rest of $2k$ be $-1$. For $n=16 \cdot k + 12$ with $k$ a nonnegative integer take : $ a_1 = a_2 = 2 , a_3 = 4k+3 $ and let $14k+7$ of the other numbers be $1$ and the rest of $2k+2$ be $-1$. By checking all possible cases , we obtain that $n=4$ is not smooth. For $n=16 \cdot k + 4$ with $k$ a positive integer take : $ a_1 = -2 , a_2 = 8k+2 $ and let $12k+3$ of the other numbers be $1$ and the rest of $4k-1$ be $-1$. Hence : The smooth numbers are the positive integers which are $0$ or $1$ mod 4 , with the exception of $4$.

MathGan wrote: Moreover , if $n$ is $16k + 4$ for a nonnegative integer $k$, $n$ cannot be smooth. Your solution is almost correct. Except that the case $n=16k+4$ is also possible for $k>1$: E.g. for $n=20$, we can take $(-1,-1,-1, -2,10,1,1,1,1,\dotsc,1)$. So the answer is: All $n \equiv 0,1 \mod 4$ except for $n=4$.

Tintarn wrote: MathGan wrote: Moreover , if $n$ is $16k + 4$ for a nonnegative integer $k$, $n$ cannot be smooth. Your solution is almost correct. Except that the case $n=16k+4$ is also possible for $k>1$: E.g. for $n=20$, we can take $(-1,-1,-1, -2,10,1,1,1,1,\dotsc,1)$. So the answer is: All $n \equiv 0,1 \mod 4$ except for $n=4$. Thanks for correcting me. I already fixed my solution.