Yes, the conjecture is true.

You can assume without loss of generality that $\varphi(1)=\rho(1)$ is smaller than the other angles. Pick $\alpha>0$ so that for all $i>1$ you have $\varphi_i>\alpha+\rho(1)$. It is also clear that it suffices to find one $n>1$ such that $\rho(n)\geq \rho(1)$.

For $N$ large enough one can assure by Dirichlet principle that there is $l$ in the range $1\leq l\leq N$ with $l\varphi_i/\pi$ within $\alpha/2$ from the nearest integer for all $i$. There are two cases to consider:

Case 1. $\{l\varphi_1/\pi\} \in [0,\alpha/2)$. In this case we can use $n=l+1$.

Case 2. $\{l\varphi_1/\pi\} \in (1-\alpha/2,1)$. This is a bit more subtle. First of all, we assume that for $l\in[1,N]$ with the above property this is the $l$ with the smallest $1-\{l\varphi_1/\pi\}$. Then we increase the range of multiples to find $m$ such that all $\{m\varphi_i/\pi\}$ are closer than $1-\{l\varphi_1/\pi\} $ to the nearest integer. If this $m$ falls into Case $1$, then we are done. Else, by our assumption on $l$ we have $m>l$ and $n=m-l+1$ works.

Yes, the conjecture is true.

You can assume without loss of generality that $\varphi(1)=\rho(1)$ is smaller than the other angles. Pick $\alpha>0$ so that for all $i>1$ you have $\varphi_i>\pi\alpha+\rho(1)$. It is also clear that it suffices to find one $n>1$ such that $\rho(n)\geq \rho(1)$.

Consider points $\{n\varphi/\pi\}$ in the hypercube $[0,1)^k$. Subdivide the hypercube into small subcubes of size less than $\alpha/2$. By Dirichlet principle, for $N$ large enough two of the points with $n\in [1,N+1]$ lie in one of the small subcubes. Their difference provides a number $l$ in the range $1\leq l\leq N$ with $l\varphi_i/\pi$ within $\alpha/2$ from the nearest integer for all $i$. There are two cases to consider:

Case 1. $\{l\varphi_1/\pi\} \in [0,\alpha/2)$. In this case we can use $n=l+1$.

Case 2. $\{l\varphi_1/\pi\} \in (1-\alpha/2,1)$. This is a bit more subtle. First of all, we assume that for $l\in[1,N]$ with the above property this is the $l$ with the smallest $1-\{l\varphi_1/\pi\}$. Then we increase the range of multiples to find $m$ such that all $\{m\varphi_i/\pi\}$ are closer than $1-\{l\varphi_1/\pi\} $ to the nearest integer. If this $m$ falls into Case $1$, then we are done. Else, by our assumption on $l$ we have $m>l$ and $n=m-l+1$ works.

Yes, the conjecture is true.

You can assume without loss of generality that $\varphi(1)=\rho(1)$ is smaller than the other angles. Pick $\alpha>0$ so that for all $i>1$ you have $\varphi_i>\alpha+\rho(1)$. It is also clear that it suffices to find one $n>1$ such that $\rho(n)\geq \rho(1)$.

For $N$ large enough one can assure by Dirichlet principle that there is $l$ in the range $1\leq l\leq N$ with $l\varphi_i/\pi$ within $\alpha/2$ from the nearest integer for all $i$. There are two cases to consider:

Case 1. $\{l\varphi_1/\pi\} \in [0,\alpha/2)$. In this case we can use $n=l+1$.

Case 2. $\{l\varphi_1/\pi\} \in (1-\alpha/2,1)$. This is a bit more subtle. First of all, we assume that for $1\in[1,N]$ this is the $l$ with the smallest $1-\{l\varphi_1/\pi\}$. Then we increase the range of multiples to find $m$ such that all $\{m\varphi_i/\pi\}$ are closer than $1-\{l\varphi_1/\pi\} $ to the nearest integer. If this $m$ falls into Case $1$, then we are done. Else, by our assumption on $l$ we have $m>l$ and $n=m-l+1$ works.

Yes, the conjecture is true.

You can assume without loss of generality that $\varphi(1)=\rho(1)$ is smaller than the other angles. Pick $\alpha>0$ so that for all $i>1$ you have $\varphi_i>\alpha+\rho(1)$. It is also clear that it suffices to find one $n>1$ such that $\rho(n)\geq \rho(1)$.

For $N$ large enough one can assure by Dirichlet principle that there is $l$ in the range $1\leq l\leq N$ with $l\varphi_i/\pi$ within $\alpha/2$ from the nearest integer for all $i$. There are two cases to consider:

Case 1. $\{l\varphi_1/\pi\} \in [0,\alpha/2)$. In this case we can use $n=l+1$.

Case 2. $\{l\varphi_1/\pi\} \in (1-\alpha/2,1)$. This is a bit more subtle. First of all, we assume that for $l\in[1,N]$ with the above property this is the $l$ with the smallest $1-\{l\varphi_1/\pi\}$. Then we increase the range of multiples to find $m$ such that all $\{m\varphi_i/\pi\}$ are closer than $1-\{l\varphi_1/\pi\} $ to the nearest integer. If this $m$ falls into Case $1$, then we are done. Else, by our assumption on $l$ we have $m>l$ and $n=m-l+1$ works.