Problem 113

Solution:
A. If $\sin x = \tan x$, then $\sin x = \tan x = \frac{\sin x}{\cos x}$,
Since $x \in (0,\pi)$, we have $\sin x \neq 0$, which implies $1 = \frac{1}{\cos x}$, i.e., $\cos x = 1$,
But since $x \in (0,\pi)$, $\cos x = 1$ does not hold. Therefore, the statement that there exists $x \in (0,\pi)$ such that $\sin x = \tan x$ is false. Hence, option A is incorrect.
B. The negation of "For any $x \in \mathbb{R}$, $x^2 + x + 1 > 0$" is "There exists $x_0 \in \mathbb{R}$ such that $x_0^2 + x_0 + 1 \leqslant 0$". Therefore, option B is incorrect.
C. When $\theta = \frac{\pi}{2}$, $f(x) = \sin(2x + \theta) = \sin(2x + \frac{\pi}{2}) = \cos 2x$ is an even function. Hence, option C is incorrect.
D. In triangle $ABC$, if $C = \frac{\pi}{2}$, then $A + B = \frac{\pi}{2}$,
Thus, from $\sin A + \sin B = \sin(\frac{\pi}{2} - B) + \sin(\frac{\pi}{2} - A) = \cos B + \cos A$, the necessity holds;
Since $\sin A + \sin B = \cos A + \cos B$,
We have $\sin A - \cos A = \cos B - \sin B$,
Squaring both sides gives $\sin^2 A - 2\sin A\cos A + \cos^2 A = \sin^2 B - 2\sin B\cos B + \cos^2 B$,
Thus, $1 - 2\sin A\cos A = 1 - 2\sin B\cos B$,
Therefore, $\sin 2A = \sin 2B$,
This implies $2A = 2B$ or $2A = \pi - 2B$,
i.e., $A = B$ or $A + B = \frac{\pi}{2}$,
When $A = B$, "$\sin A + \sin B = \cos A + \cos B$" is equivalent to $2\sin A = 2\cos A$,
Thus, $\tan A = 1$, i.e., $A = B = \frac{\pi}{4}$, and in this case, $C = \frac{\pi}{2}$,
In summary, $C = \frac{\pi}{2}$ always holds, which means the sufficiency is established,
Overall, in triangle $ABC$, "$\sin A + \sin B = \cos A + \cos B$" is a necessary and sufficient condition for "$C = \frac{\pi}{2}$". Therefore, option D is correct.
Hence, the choice is: $\boxed{D}$

A. Judged based on the properties of trigonometric functions.
B. Judged based on the negation of a universal proposition being an existential proposition.
C. Judged based on the odd-even properties of trigonometric functions.
D. Judged based on the definitions of necessary and sufficient conditions, using the squaring method.
This question mainly examines the judgment of the truthfulness of propositions, involving a wide range of knowledge points and requiring strong comprehensive skills to test students' calculation and reasoning abilities.