IRL ABooklet

Let $a < b < c < d < e$ be the elements of $A$. The two smallest sums are $25 = a + b < 31 = a + c$. The two largest are $57 = c + e < 71 = d + e$. We then obtain
$$a + d = (a + c) - (c + e) + (d + e) = 31 - 57 + 71 = 45$$
$$e - a = (c + e) - (a + c) = 57 - 31 = 26.$$

This allows us to express the elements of $A$ in terms of $a$ as follows:
$$e = 26 + a, \quad d = 45 - a, \quad c = 31 - a, \quad b = 25 - a.$$
For $a < b = 25 - a$ we need $2a < 25$, i.e. $a \le 12$. For $d < e$ we need $45 - a < 26 + a$ or $19 < 2a$, i.e. $10 \le a$. Therefore we only have the following three possibilities:

<table>
<thead>
<tr><th>a</th><th>10</th><th>11</th><th>12</th></tr>
</thead>
<tbody>
<tr><th>b = 25 - a</th><td>15</td><td>14</td><td>13</td></tr>
<tr><th>c = 31 - a</th><td>21</td><td>20</td><td>19</td></tr>
<tr><th>d = 45 - a</th><td>35</td><td>34</td><td>33</td></tr>
<tr><th>e = 26 + a</th><td>36</td><td>37</td><td>38</td></tr>
</tbody>
</table>

A straightforward check shows that these three possibilities indeed satisfy the given conditions:

<table>
<thead>
<tr><th>A</th><th>S</th></tr>
</thead>
<tbody>
<tr><td>10, 15, 21, 35, 36</td><td>25, 31, 36, 45, 46, 50, 51, 56, 57, 71</td></tr>
<tr><td>11, 14, 20, 34, 37</td><td>25, 31, 34, 45, 48, 51, 54, 57, 71</td></tr>
<tr><td>12, 13, 19, 33, 38</td><td>25, 31, 32, 45, 46, 50, 51, 52, 57, 71</td></tr>
</tbody>
</table>

The set $S$ in the second case has only 9 elements because $48 = 11 + 37 = 14 + 34$.