Let's write the following sum in a simpler form:

$$
S=\sum_{i=1}^{n} \frac{n-i}{(2 i+1)(n+i+1)} .
$$
Notice that in the general term of the sum - that is, in every term - the numerator is equal to the difference of the second and first factors of the denominator. Therefore, each term can be written as the difference of two fractions, both of which have a numerator of 1, and denominators that are the said factors:

$$
\frac{n-i}{(2 i+1)(n+i+1)}=\frac{(n+i+1)-(2 i+1)}{(2 i+1)(n+i+1)}=\frac{1}{2 i+1}-\frac{1}{n+i+1}
$$

Based on this, $S$ can be written as the difference of two other sums:

$$
S=\sum_{i=1}^{n} \frac{1}{2 i+1}-\sum_{i=1}^{n} \frac{1}{n+i+1}=S_{1}-S_{2}
$$

The denominators in the two new sums:

$$
3,5,7, \ldots, 2 n+1, \quad \text { and } \quad n+2, n+3, \ldots, 2 n+1
$$

increase by twos and ones, respectively, and the largest term in both sequences is the same. We can also say: among the natural numbers from 1 to $2 n+1$, the first one and the even numbers are missing in the first sequence, while in the second sequence, every natural number up to $n+1$ is missing. Based on this, both sums can be written as differences: taking as the minuend the sum of the reciprocals of the natural numbers up to $2 n+1$ - which will thus be common - and as the subtrahend the sum of the reciprocals of the numbers thus introduced, which did not originally appear:

$$
S_{1}=\sum_{k=1}^{2 n+1} \frac{1}{k}-\left(1+\sum_{k=1}^{n} \frac{1}{2 k}\right)=\sum_{k=1}^{2 n+1} \frac{1}{k}-1-\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k}, \quad S_{2}=\sum_{k=1}^{2 n+1} \frac{1}{k}-\sum_{k=1}^{n+1} \frac{1}{k}
$$

The common term disappears from their difference, and by writing the last term of the first sum separately, simplification is possible:

$$
S_{1}=\sum_{k=1}^{n+1} \frac{1}{k}-\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k}-1=\frac{1}{n+1}+\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k}-1=\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k}-\frac{n}{n+1}
$$

which can be considered simpler compared to the original form. The sum appearing here cannot be written in closed form as a function of $n$.

Szendrei Ágnes (Szeged, Ságvári E. Gymnasium, 4th grade)