Let me begin by restating your conjecture. Consider the sum $$S(x):=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$ Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $b_j=0$ for all even $j$ (in particular, $n$ is odd).

First of all, notice that $S(x)\geq 0$ on $[0,\pi]$ if and only if $S^*(x)\geq 0$ on $[-\pi,\pi]$, where $$S^*(x):=2S(x)\sin x=\sum_{j=1}b_j\left(\cos(j-1)x-\cos(j+1)x\right)\geq 0.$$

Second, we must have $b_1=1$, by the well-known theorem that a trigonometric polynomial without a constant term must change sign on $[-\pi,\pi]$. (This is because its integral over $[-\pi,\pi]$ equals zero).

Now transform $S^*$ as follows: $$S^*(z)=1-\cos(n+1)x+A(x)+B(x),$$ where $$A(x)=b_2\cos x+(b_3-1)\cos 2x+(1-b_{n-2})\cos(n-1)x-b_{n-1}\cos nx,$$ and $$B(x)=\sum_{j=3}^{n-2}(b_{j+1}-b_{j-1})\cos jx.$$ Let us set $z=\exp(ix),\;|z|=1$ and consider the first two summands in $S^*$: $$1-\cos(n+1)x=(2-z^{n+1}-z^{-n-1})/2.$$ This is non-negative and has double zeros at the roots $z_k$ of degree $n+1$ of unity. So for $S^*$ to be non-negative, it is necessary that $A+B$ be non-negative at the points $x_k=2\pi k/(n+1)$ corresponding to $z_k$. On the other hand, we have $$\sum_{k=0}^n A(x_k)+B(x_k)=0, \quad x_k=2\pi k/(n+1),$$ by the well-known "orthogonality relations", $$\sum_k z_k^m=0,\quad\mbox{when}\quad |m|\leq n.$$ Therefore $A+B$ must be zero at all $n+1$-st roots of unity, moreover, all these zeros must be multiple (if not, $S^*(x)$ will change sign near some $x_k$), from which we conclude that $A+B\equiv 0$, because a non-zero trigonometric polynomial of degree $n$ cannot have $n+1$ multiple zeros. This means that $$b_2=0,\; b_3=1,\; b_{n-2}=1,\; b_{n-1}=0,$$ and $b_{j+1}=b_{j-1}$ for all $j\in[3,n-2]$. This proves your statement.

Let me begin by restating your conjecture. Consider the sum $$S(x):=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$ Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $b_j=0$ for all even $j$ (in particular, $n$ is odd).

First of all, notice that $S(x)\geq 0$ on $[0,\pi]$ if and only if $S^*(x)\geq 0$ on $[-\pi,\pi]$, where $$S^*(x):=2S(x)\sin x=\sum_{j=1}^nb_j\left(\cos(j-1)x-\cos(j+1)x\right)\geq 0.$$

Second, we must have $b_1=1$, by the well-known theorem that a trigonometric polynomial without a constant term must change sign on $[-\pi,\pi]$. (This is because its integral over $[-\pi,\pi]$ equals zero).

Now transform $S^*$ as follows: $$S^*(z)=1-\cos(n+1)x+A(x)+B(x),$$ where $$A(x)=b_2\cos x+(b_3-1)\cos 2x+(1-b_{n-2})\cos(n-1)x-b_{n-1}\cos nx,$$ and $$B(x)=\sum_{j=3}^{n-2}(b_{j+1}-b_{j-1})\cos jx.$$ Let us set $z=\exp(ix),\;|z|=1$ and consider the first two summands in $S^*$: $$1-\cos(n+1)x=(2-z^{n+1}-z^{-n-1})/2.$$ This is non-negative and has double zeros at the roots $z_k$ of degree $n+1$ of unity. So for $S^*$ to be non-negative, it is necessary that $A+B$ be non-negative at the points $x_k=2\pi k/(n+1)$ corresponding to $z_k$. On the other hand, we have $$\sum_{k=0}^n \left(A(x_k)+B(x_k)\right)=0, \quad x_k=2\pi k/(n+1),$$ by the well-known "orthogonality relations", $$\sum_{k=0}^n z_k^m=0,\quad\mbox{when}\quad |m|\leq n.$$ Therefore $A+B$ must be zero at all $n+1$-st roots of unity, moreover, all these zeros must be multiple (if not, $S^*(x)$ will change sign near some $x_k$), from which we conclude that $A+B\equiv 0$, because a non-zero trigonometric polynomial of degree $n$ cannot have $n+1$ multiple zeros. This means that $$b_2=0,\; b_3=1,\; b_{n-2}=1,\; b_{n-1}=0,$$ and $b_{j+1}=b_{j-1}$ for all $j\in[3,n-2]$. This proves your statement.

Let me begin by restating your conjecture. Consider the sum $$S=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$ Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $b_j=0$ for all even $j$ (in particular, $n$ is odd).

First of all, notice that $S(x)\geq 0$ on $[0,\pi]$ if and only if $$S^*(x):=2S(x)\sin x=\sum_{j=1}b_j\left(\cos(j-1)x-\cos(j+1)x\right)\geq 0$$ on $[-\pi,\pi]$.

Second, we must have $b_1=1$, by the well-known theorem that a trigonometric polynomial without a constant term must change sign on $[-\pi,\pi]$. (This is because its integral over $[-\pi,\pi]$ equals zero).

Now transform $S^*$ as follows: $$S^*(z)=1-\cos(n+1)x+A(x)+B(x),$$ where $$A(x)=b_2\cos x+(b_3-1)\cos 2x+(1-b_{n-2})\cos(n-1)x-b_{n-1}\cos nx,$$ and $$B(x)=\sum_{j=3}^{n-2}(b_{j+1}-b_{j-1})\cos jx.$$ Let us set $z=\exp(ix),\;|z|=1$ and consider the first two summands in $S^*$: $$1-\cos(n+1)x=(2-z^{n+1}-z^{-n-1})/2.$$ This is non-negative and has double zeros at the roots $z_k$ of degree $n+1$ of unity. So for $S^*$ to be non-negative, it is necessary that $A+B$ be non-negative at the points corresponding to all $z_k$. On the other hand, we have $$\sum_{k=0}^n A(x_k)+B(x_k)=0, \quad x_k=2\pi k/(n+1),$$ by the well-known "orthogonality relations", $$\sum_k z_k^m=0,\quad\mbox{when}\quad |m|\leq n.$$ Therefore $A+B$ must be zero at all $n+1$-st roots of unity, moreover, all these zeros must be multiple, from which we conclude that $A+B\equiv 0$. This means that $$b_2=0,\; b_3=1,\; b_{n-2}=1,\; b_{n-1}=0,$$ and $b_{j+1}=b_{j-1}$ for all $j\in[3,n-2]$. This proves your statement.

Let me begin by restating your conjecture. Consider the sum $$S(x):=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$ Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $b_j=0$ for all even $j$ (in particular, $n$ is odd).

First of all, notice that $S(x)\geq 0$ on $[0,\pi]$ if and only if $S^*(x)\geq 0$ on $[-\pi,\pi]$, where $$S^*(x):=2S(x)\sin x=\sum_{j=1}b_j\left(\cos(j-1)x-\cos(j+1)x\right)\geq 0.$$

Second, we must have $b_1=1$, by the well-known theorem that a trigonometric polynomial without a constant term must change sign on $[-\pi,\pi]$. (This is because its integral over $[-\pi,\pi]$ equals zero).

Now transform $S^*$ as follows: $$S^*(z)=1-\cos(n+1)x+A(x)+B(x),$$ where $$A(x)=b_2\cos x+(b_3-1)\cos 2x+(1-b_{n-2})\cos(n-1)x-b_{n-1}\cos nx,$$ and $$B(x)=\sum_{j=3}^{n-2}(b_{j+1}-b_{j-1})\cos jx.$$ Let us set $z=\exp(ix),\;|z|=1$ and consider the first two summands in $S^*$: $$1-\cos(n+1)x=(2-z^{n+1}-z^{-n-1})/2.$$ This is non-negative and has double zeros at the roots $z_k$ of degree $n+1$ of unity. So for $S^*$ to be non-negative, it is necessary that $A+B$ be non-negative at the points $x_k=2\pi k/(n+1)$ corresponding to $z_k$. On the other hand, we have $$\sum_{k=0}^n A(x_k)+B(x_k)=0, \quad x_k=2\pi k/(n+1),$$ by the well-known "orthogonality relations", $$\sum_k z_k^m=0,\quad\mbox{when}\quad |m|\leq n.$$ Therefore $A+B$ must be zero at all $n+1$-st roots of unity, moreover, all these zeros must be multiple (if not, $S^*(x)$ will change sign near some $x_k$), from which we conclude that $A+B\equiv 0$, because a non-zero trigonometric polynomial of degree $n$ cannot have $n+1$ multiple zeros. This means that $$b_2=0,\; b_3=1,\; b_{n-2}=1,\; b_{n-1}=0,$$ and $b_{j+1}=b_{j-1}$ for all $j\in[3,n-2]$. This proves your statement.