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Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had earlier deleted, has been edited in response to some helpful comments.)

If $m$ is odd and square-free, then the claim of the OP holds.

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) = 0.$$

Here $N = 2m$ is square-free, and in that case the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$ (see this mathstackexchange discussion: http://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).

For $N = 2m$, any odd $k \in S$ is prime to $N$ and hence $e^{k \pi i/m}, e^{-k\pi i/m}$ are primitive $N$-th roots of unity. If $k \in S$ is even, then $m + k$ is odd and prime to $m$ and thus to $N$, so $e^{(m + k)\pi i/m} = -e^{k\pi i/m}$ is also primitive $N$-th root of unity, as is its conjugate $e^{(m - k)\pi i/m}$; notice the $m-k$ lie in $\{j: m/2 \leq j \leq m, \gcd(j,m) = 1\}$ which is disjoint from $S$. Then

$$\begin{array}{lll} \sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) \\ & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{(m-k)\pi i/m} - e^{-(m-k)\pi i/m}) \end{array}$$

where all the primitive roots of unity appearing in the last expression are manifestly distinct. By linear independence of the primitive roots, if that linear combination is zero, then $a_k = 0$ for all $k$, as required.

Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had earlier deleted, has been edited in response to some helpful comments.)

If $m$ is odd and square-free, then the claim of the OP holds.

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) = 0.$$

Here $N = 2m$ is square-free, and in that case the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$ (see this mathstackexchange discussion: https://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).

For $N = 2m$, any odd $k \in S$ is prime to $N$ and hence $e^{k \pi i/m}, e^{-k\pi i/m}$ are primitive $N$-th roots of unity. If $k \in S$ is even, then $m + k$ is odd and prime to $m$ and thus to $N$, so $e^{(m + k)\pi i/m} = -e^{k\pi i/m}$ is also primitive $N$-th root of unity, as is its conjugate $e^{(m - k)\pi i/m}$; notice the $m-k$ lie in $\{j: m/2 \leq j \leq m, \gcd(j,m) = 1\}$ which is disjoint from $S$. Then

$$\begin{array}{lll} \sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) \\ & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{(m-k)\pi i/m} - e^{-(m-k)\pi i/m}) \end{array}$$

where all the primitive roots of unity appearing in the last expression are manifestly distinct. By linear independence of the primitive roots, if that linear combination is zero, then $a_k = 0$ for all $k$, as required.

Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had deleted, has been edited in response to some helpful comments.)

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(\exp(k\pi i/m) - \exp(-k\pi i/m)) = 0$$

where the $\exp(k\pi i/m), \exp(-k\pi i/m)$ are manifestly distinct primitive $(2m)$-th roots of unity.

In case $N = 2m$ is square-free, then according to the mathstackexchange discussion http://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension, the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$. This would clearly force all the $a_k$ to be $0$.

Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had earlier deleted, has been edited in response to some helpful comments.)

If $m$ is odd and square-free, then the claim of the OP holds.

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) = 0.$$

Here $N = 2m$ is square-free, and in that case the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$ (see this mathstackexchange discussion: http://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).

For $N = 2m$, any odd $k \in S$ is prime to $N$ and hence $e^{k \pi i/m}, e^{-k\pi i/m}$ are primitive $N$-th roots of unity. If $k \in S$ is even, then $m + k$ is odd and prime to $m$ and thus to $N$, so $e^{(m + k)\pi i/m} = -e^{k\pi i/m}$ is also primitive $N$-th root of unity, as is its conjugate $e^{(m - k)\pi i/m}$; notice the $m-k$ lie in $\{j: m/2 \leq j \leq m, \gcd(j,m) = 1\}$ which is disjoint from $S$. Then

$$\begin{array}{lll} \sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) \\ & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{(m-k)\pi i/m} - e^{-(m-k)\pi i/m}) \end{array}$$

where all the primitive roots of unity appearing in the last expression are manifestly distinct. By linear independence of the primitive roots, if that linear combination is zero, then $a_k = 0$ for all $k$, as required.

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(\exp(k\pi i/m) - \exp(-k\pi i/m)) = 0$$

where the $\exp(k\pi i/m), \exp(-k\pi i/m)$ are manifestly distinct primitive $(2m)$-th roots of unity. But it is well-known that for any $N$, the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$. See for example Lang's Algebra, Theorem VI.3.1. This forces the $a_k$ to be $0$.

Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had deleted, has been edited in response to some helpful comments.)

Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then

$$\sum_{k \in S} a_k(\exp(k\pi i/m) - \exp(-k\pi i/m)) = 0$$

where the $\exp(k\pi i/m), \exp(-k\pi i/m)$ are manifestly distinct primitive $(2m)$-th roots of unity.

In case $N = 2m$ is square-free, then according to the mathstackexchange discussion http://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension, the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$. This would clearly force all the $a_k$ to be $0$.