10. Given $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ are all non-negative, and $\theta_{1}+\theta_{2}+\cdots+\theta_{n}=\pi$, find the maximum value of $\sin ^{2} \theta_{1}+$ $\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{n}$. (1985 IMO Shortlist)
10. First consider the case where $\theta_{1}+\theta_{2}$ is a constant. In this case,
$$\begin{array}{l}
\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}=\left(\sin \theta_{1}+\sin \theta_{2}\right)^{2}-2 \sin \theta_{1} \sin \theta_{2}= \\
4 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2} \cos ^{2} \frac{\theta_{1}-\theta_{2}}{2}-\cos \left(\theta_{1}-\theta_{2}\right)+\cos \left(\theta_{1}+\theta_{2}\right)= \\
2 \cos ^{2} \frac{\theta_{1}-\theta_{2}}{2}\left(2 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2}-1\right)+1+\cos \left(\theta_{1}+\theta_{2}\right)
\end{array}$$

The last two terms and the factor in the first term's parentheses on the right side of the equation are constants, and we have
$$2 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2}-1\left\{\begin{array}{l}
0, \text { when } \theta_{1}+\theta_{2}>\frac{\pi}{2}
\end{array}\right.$$

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 (1).

When $n \geqslant 4$, there are always two angles whose sum does not exceed $\frac{\pi}{2}$. Hence, one of these angles can be set to 0, and the other to the sum of the original two angles, without decreasing the sum of the squares of the sines. This way, the sum of the squares of the sines of $n$ angles can be reduced to the case of three angles.

Assume $n=3$. If $\theta_{1}, \theta_{2}, \theta_{3}$ include two angles equal to $\frac{\pi}{2}$ and one angle equal to 0, then these can be changed to $\frac{\pi}{2}$, $\frac{\pi}{4}$, $\frac{\pi}{4}$.

Assume $\theta_{1} \leqslant \theta_{2} \leqslant \theta_{3}, \theta_{1}<\frac{\pi}{2}, \theta_{1}<\frac{\pi}{3}<\theta_{3}$. Let $\theta_{1}^{\prime}=\frac{\pi}{3}$, $\theta_{2}^{\prime}=\theta_{2}$, $\theta_{3}^{\prime}=\theta_{1}+\theta_{3}-\theta_{1}^{\prime}$, then $\theta_{1}^{\prime}+\theta_{3}^{\prime}=\theta_{1}+\theta_{3},\left|\theta_{1}^{\prime}-\theta_{3}^{\prime}\right|<\left|\theta_{1}-\theta_{3}\right|$.

Therefore, by (1) and (2),
$$\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\sin ^{2} \theta_{3}<\sin ^{2} \theta^{\prime}{ }_{1}+\sin ^{2} \theta^{\prime}{ }_{2}+\sin ^{2} \theta_{3}^{\prime}$$

Since $\theta^{\prime}{ }_{2}+\theta^{\prime}{ }_{3}=\pi-\theta^{\prime}{ }_{1}=\frac{2 \pi}{3}$, by (1) and (2) again,
$$\sin ^{2} \theta_{1}^{\prime}+\sin ^{2} \theta_{3}^{\prime} \leqslant 2 \sin ^{2} \frac{\pi}{3}=\frac{3}{2}$$

Thus,
$$\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\sin ^{2} \theta_{3} \leqslant \frac{9}{4}$$

The 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 $n=2$, since $\theta_{1}+\theta_{2}=\pi$, we have $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}=2 \sin ^{2} \theta_{1} \leqslant 2$, so the maximum value is 2.