3. First, consider when $\theta_{1}+\theta_{2}$ is constant, we have $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}=\left(\sin \theta_{1}+\sin \theta_{2}\right)^{2}-$ $2 \sin \theta_{1} \sin \theta_{2}=4 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2} \cos ^{2} \frac{\theta_{1}-\theta_{2}}{2}-\cos \left(\theta_{1}-\theta_{2}\right)+\cos \left(\theta_{1}+\theta_{2}\right)=$ $2 \cos ^{2} \frac{\theta_{1}-\theta_{2}}{2}\left(2 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2}-1\right)+1+\cos \left(\theta_{1}+\theta_{2}\right)$, therefore, when $\theta_{1}+\theta_{2}\frac{\pi}{2}$, the smaller the absolute value of the difference between the two angles, the larger the value of the above expression.

When $n \geqslant 4$, there are always two angles such that their sum $\leqslant \frac{\pi}{2}$. These two angles can be adjusted to one being 0 and the other being the sum of the original two angles without reducing the sum of the squares of the sines. In this way, the sum of the squares of the sines of $n$ angles is reduced to the case of three angles.

When $n=3$, if $\theta_{1} 、 \theta_{2} 、 \theta_{3}$ include two angles of $\frac{\pi}{2}$ and one angle of 0, they can be changed to $\frac{\pi}{2} 、 \frac{\pi}{4} 、 \frac{\pi}{4}$. Let $\theta_{1} \leqslant \theta_{2} \leqslant \theta_{3}, \theta_{1}\frac{\pi}{2}, \theta_{1}<\frac{\pi}{3}<\theta_{3}$.

Let $\theta_{1}^{\prime}=\frac{\pi}{3}, \theta_{2}^{\prime}=\theta_{2}, \theta_{3}^{\prime}=\theta_{1}+\theta_{3}-\theta_{1}^{\prime}$, then $\theta_{1}^{\prime}+\theta_{3}^{\prime}=\theta_{1}+\theta_{3},\left|\theta_{1}^{\prime}-\theta_{3}^{\prime}\right|<$ $\left|\theta_{1}-\theta_{3}\right|$, from the previous discussion, we know $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\sin ^{2} \theta_{3}<\sin ^{2} \theta_{1}^{\prime}+\sin ^{2} \theta_{2}^{\prime}+\sin ^{2} \theta_{3}^{\prime}$. Since $\theta^{\prime}{ }_{2}+\theta_{3}^{\prime}=\frac{2 \pi}{3}$, we also get $\sin ^{2} \theta_{2}^{\prime}+\sin ^{2} \theta_{3}^{\prime} \leqslant 2 \sin ^{2} \frac{\pi}{3}=\frac{3}{2}$, hence $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+$ $\sin ^{2} \theta_{3} \leqslant \frac{9}{4}$, equality holds if and only if $\theta_{1}=\theta_{2}=\theta_{3}=\frac{\pi}{3}$. Therefore, when $n \geqslant 3$, the maximum value sought is $\frac{9}{4}$.

When $n=2$, $\theta_{1}+\theta_{2}=\pi$, thus $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}=2 \sin ^{2} \theta_{1} \leqslant 2$. Therefore, the maximum value is 2.
3 Given $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ are all non-negative, and $\theta_{1}+\theta_{2}+\cdots+\theta_{n}=\pi$. Find
$$\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{n}$$

the maximum value.