I give a way that may work for odd numbers. This is too long for a comment.
First, the quantity $$\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big) = 2\cos\Big(\pi\Big(\frac{m}{p}+\frac{n}{q}\Big)\Big) cos\Big(\pi\Big(\frac{m}{p}-\frac{n}{q}\Big)\Big)$$$$\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big) = 2\cos\Big(\pi\Big(\frac{m}{p}+\frac{n}{q}\Big)\Big) \cos\Big(\pi\Big(\frac{m}{p}-\frac{n}{q}\Big)\Big)$$ cannot vanish since $p$ and $q$ are coprime. Hence the sum is well-defined.
Call $T_p$ the TchebychevChebyshev polynomial, defined by the equality
$T_p(\cos(\theta)) = \cos(p\theta)$ for all real numbernumbers $\theta$. Then for all integerintegers $m$,
$$T_p\Big(\cos\Big(\frac{2m\pi}{p}\Big)\Big) = T_p(\cos(2m\pi)) = T_p(\cos(0)) = 1.$$
If $p$ is odd, the real numbers $\cos(2m\pi/p)$ for $0 \le m \le (p-1)/2$ are distinct roots of $T_p$. The multiplicity is $2$ if $1 \le m \le (p-1)/2$, and $1$ if $m=0$. Therefore, we have all roots. Since $T_p$ has leading term $2^{p-1}X^p$, we derive
$$T_p - 1 = 2^{p-1}(X-1)\prod_{m=1}^{(p-1)/2}\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)\Big) = 2^{p-1} \prod_{m=1}^p\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)\Big).$$
If $q$ is odd, a similar formula holds for $q$.
Given an integer $n \in [1,q]$, one has
$$T_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1
= 2^{p-1} \prod_{m=1}^p\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)\Big).$$
Thus
$$\prod_{n=1}^q\Big( T_p \Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1 \Big)
= 2^{(p-1)q} \prod_{m=1}^p \prod_{n=1}^q \Big(X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)\Big).$$
Call $c_{pq}X^{pq} + \cdots + c_0$ this polynomial. The sum to be estimated is the sum of the reciprocal of its roots, which is $-c_1/c_0$. So we are lead to estimate the coefficients of this polynomial...
Another strategy is to write for each integer $n \in [1,q]$, $$\frac{T'_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{T_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1} = \sum_{m=1}^p\frac{1}{X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)},$$ so $$\sum_{m=1}^p\frac{1}{\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big)} = \frac{T'_p\Big(-\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{1 - T_p\Big(-\cos\Big(\frac{2n\pi}{q}\Big)\Big)} = \frac{T'_p\Big(\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{1 + T_p\Big(\cos\Big(\frac{2n\pi}{q}\Big)\Big)},$$ since $T_p$ is an odd function.
Using the equalities $T_p(\cos(\theta)) = \cos(p\theta)$ and $-T'_p(\cos(\theta))\sin\theta = - p \sin(p\theta)$, one derives $$\sum_{m=1}^p\frac{1}{\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big)} = p \frac{\sin(2np\pi/q)}{\sin(2n\pi/q)} \times \frac{1}{1+\cos(2np\pi/q)}.$$ It remains to sum over all integers $n \in [1,q]$ and to bound above...
