China National Olympiad

Let $a_i, b_i$ for $i = 1, \ldots, n$ be nonnegative numbers, and let $n \geq 4$ such that $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i > 0$.

We aim to find the maximum value of the expression:
$$\frac{\sum_{i=1}^n a_i(a_i + b_i)}{\sum_{i=1}^n b_i(a_i + b_i)}.$$

We will prove that for $n \geq 4$, the maximum value is $n - 1$.

Without loss of generality, we can assume $a_1 \geq a_2 \geq \cdots \geq a_n$ and $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i = 1$. Let $A = \sum_{i=1}^n a_i^2$, $B = \sum_{i=1}^n b_i^2$, and $X = \sum_{i=1}^n a_i b_i$.

We define the function:
$$f(X) = \frac{A + X}{B + X}.$$

The derivative of $f(X)$ is:
$$f'(X) = \frac{B - A}{(B + X)^2}.$$

Since $f'(X) < 0$ when $A > B$, we want to minimize $X$ to maximize the fraction. By the rearrangement inequality, $X$ is minimized when $b_1 \leq b_2 \leq \cdots \leq b_n$.

Next, we fix $b_1, \ldots, b_n$ and define:
$$F(a_1, \ldots, a_n) = \frac{A + X}{B + X}.$$

We will show that:
$$F(a_1, a_2, \ldots, a_n) \leq F(a_1 + a_2, 0, \ldots, a_n).$$

This is true because $a_1^2 + a_2^2 \leq (a_1 + a_2)^2$ and $a_1 b_1 + a_2 b_2 \geq (a_1 + a_2) b_1$. By repeating this step $n - 1$ times, we conclude:
$$F(a_1, \ldots, a_n) \leq F(1, 0, \ldots, 0).$$

It remains to prove:
$$F(1, 0, \ldots, 0) = \frac{1 + b_1}{\sum_{i=1}^n b_i^2 + b_1} \leq n - 1.$$

Using the Cauchy-Schwarz inequality, we get:
$$\frac{1 + b_1}{\sum_{i=1}^n b_i^2 + b_1} \leq \frac{1 + b_1}{b_1^2 + \frac{(1 - b_1)^2}{n - 1} + b_1}.$$

Finally, we need to show:
$$\frac{1 + b_1}{b_1^2 + \frac{(1 - b_1)^2}{n - 1} + b_1} \leq n - 1 \quad \Leftrightarrow \quad b_1 (n b_1 + n - 4) \geq 0.$$

This inequality clearly holds for $n \geq 4$.

Equality holds when $a_1 = 1$, $a_2 = \cdots = a_n = 0$, and $b_1 = 0$, $b_2 = \cdots = b_n = \frac{1}{n - 1}$.

Thus, the maximum value of the given expression is:
$$\frac{\sum_{i=1}^n a_i(a_i + b_i)}{\sum_{i=1}^n b_i(a_i + b_i)} = n - 1.$$

The answer is: $\boxed{n}$ - 1}.