A $39$-tuple of real numbers $(x_1,x_2,\ldots x_{39})$ satisfies
\[2\sum_{i=1}^{39} \sin(x_i) = \sum_{i=1}^{39} \cos(x_i) = -34.\] 
The ratio between the maximum of $\cos(x_1)$ and the maximum of $\sin(x_1)$ over all tuples $(x_1,x_2,\ldots x_{39})$ satisfying the condition is $\tfrac ab$ for coprime positive integers $a$, $b$ (these maxima aren't necessarily achieved using the same tuple of real numbers). Find $a + b$.

[i]Proposed by Evan Chang[/i]
1. **Modeling the Problem:**
   We start by modeling each \( x_i \) as a vector \( \vec{v_i} = (\cos(x_i), \sin(x_i)) \) with unit magnitude. Given the conditions:
   \[
   2\sum_{i=1}^{39} \sin(x_i) = -34 \quad \text{and} \quad \sum_{i=1}^{39} \cos(x_i) = -34,
   \]
   we can rewrite these as:
   \[
   \sum_{i=1}^{39} \sin(x_i) = -17 \quad \text{and} \quad \sum_{i=1}^{39} \cos(x_i) = -34.
   \]
   This implies that the sum of all vectors \( \vec{v_i} \) is \( (-34, -17) \), which we will call point \( Q \).

2. **Geometric Interpretation:**
   We need to find the maximum values of \( \cos(x_1) \) and \( \sin(x_1) \) over all tuples \( (x_1, x_2, \ldots, x_{39}) \) satisfying the given conditions. To do this, we consider a point \( P \) on the unit circle where \( \vec{v_1} = \vec{OP} \), and we need to maximize the \( x \)-coordinate (for \( \cos(x_1) \)) and the \( y \)-coordinate (for \( \sin(x_1) \)) of \( P \).

3. **Distance Calculation:**
   The distance \( PQ \) must be less than or equal to 38 because the sum of the magnitudes of the remaining 38 vectors must be 38 (since each vector has a unit magnitude). Therefore, we set \( PQ = 38 \) exactly to maximize the coordinates.

4. **Intersection Points:**
   We look at the intersections of the circle with radius 38 centered at \( Q \) with the unit circle. These intersections will give us the maximum possible values for \( \cos(x_1) \) and \( \sin(x_1) \).

5. **Trigonometric Calculations:**
   We need to find the coordinates of these intersection points. Let \( \alpha \) be the angle such that:
   \[
   \sin(\alpha) = \frac{38}{\sqrt{38^2 + 1^2}} = \frac{38}{\sqrt{1445}} = \frac{38}{17\sqrt{5}},
   \]
   and
   \[
   \cos(\alpha) = \frac{1}{\sqrt{38^2 + 1^2}} = \frac{1}{17\sqrt{5}}.
   \]

6. **Maximizing \( \cos(x_1) \) and \( \sin(x_1) \):**
   Using the trigonometric addition formulas, we find the \( x \)-coordinate of \( X \) (the point with the maximum \( \cos(x_1) \)) and the \( y \)-coordinate of \( Y \) (the point with the maximum \( \sin(x_1) \)):
   \[
   \cos(x_1) = \frac{38}{17\sqrt{5}} \cdot \frac{1}{\sqrt{5}} - \frac{1}{17\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{36}{85},
   \]
   and
   \[
   \sin(x_1) = \frac{38}{17\sqrt{5}} \cdot \frac{2}{\sqrt{5}} - \frac{1}{17\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{15}{17}.
   \]

7. **Ratio Calculation:**
   The ratio between the maximum of \( \cos(x_1) \) and the maximum of \( \sin(x_1) \) is:
   \[
   \frac{\frac{36}{85}}{\frac{15}{17}} = \frac{36}{85} \cdot \frac{17}{15} = \frac{36 \cdot 17}{85 \cdot 15} = \frac{612}{1275} = \frac{12}{25}.
   \]

8. **Final Answer:**
   The ratio is \( \frac{12}{25} \), and since \( a = 12 \) and \( b = 25 \), we have \( a + b = 37 \).

The final answer is \( \boxed{37} \).