2.35 Consider a set of $n$ distinct weights, where each weight is an integer not exceeding 21 grams. For what minimum value of $n$ will the set of weights necessarily contain two pairs of weights with equal sums?

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[Solution] When the weights of the 7 weights are 1, 2, 3, 5, 8, 13, 21 grams respectively, it is easy to verify that the sums of the weights of any two pairs are all different. In fact, among the 4 weights of two pairs, the sum of the weights of the pair containing the heaviest weight is heavier. From this, we know that the minimum value sought is no less than 8.

Next, we will prove that in any 8 weights, we can always find two pairs of weights such that their sums are equal.

Since 8 weights can form $C_{8}^{2}=28$ different pairs of weights (two "pairs" are different if at least one weight in one pair does not belong to the other), and the absolute value of the difference in weight (simply called the difference) of each pair of weights does not exceed 20 grams and is a positive integer, there are at most 20 different difference values. Therefore, we can select 8 different pairs of weights:
$$
\left\{\left(a_{i}, b_{i}\right),\left(c_{i}, d_{i}\right)\right\}, i=1,2, \cdots, 8 \text {, }
$$

such that
$$
b_{i}-a_{i}=d_{i}-c_{i}, i=1,2, \cdots, 8 .
$$
(1) If among the 8 pairs of weights in (1), there is a group of 4 weights $a, b, c, d$ that are all different, then by (2) we get $a+d=b+c$, and the two pairs of weights $(a, d),(b, c)$ meet the requirements.
(2) Suppose that in the 8 pairs of weights in (1), the 4 weights in each group are not all different. At this point, if there are two weights among the 8 weights that are the same, then there are obviously two pairs of weights with equal sums. Below, assume that the weights of the 8 weights are all different. Then by (2) we know $b_{i}=c_{i}, i=1,2, \cdots$, 8. In this way, each pair of weights in (1) is transformed into a triplet
$$
\left\{a_{i}, b_{i}, d_{i}\right\}, i=1,2, \cdots, 8,
$$

satisfying the condition
$$
a_{i}+d_{i}=2 b_{i}, i=1,2, \cdots, 8,
$$

that is, $b_{i}$ is the arithmetic mean of $a_{i}$ and $d_{i}$, so it cannot be the lightest or heaviest weight, but can only be one of the 6 middle weights. By the pigeonhole principle, $b_{1}, b_{2}, \cdots, b_{8}$ must have two that are the same, say $b_{i}=b_{j}(i \neq j)$. Then by (4) we have $a_{i}+d_{i}=a_{j}+d_{j}$. It is easy to see that $a_{i}, d_{i}, a_{j}$, $d_{j}$ are all different. Thus $\left(a_{i}, d_{i}\right)$ and $\left(a_{j}, d_{j}\right)$ meet the requirements of the problem.
In summary, the minimum value of the number of weights $n$ is 8.