Obviously, if \(\{a, b, c, d\}\) is a set with the property mentioned in the statement, then the set \(\{n^2a, n^2b, n^2c, n^2d\}\) is, for any natural number \(n\), a set with the desired property. Therefore, it is sufficient to find such a set. How can we find such a set? If \(a + b + c = x^2\), \(a + b + d = y^2\), \(a + c + d = z^2\), \(b + c + d = t^2\), with \(x, y, z, t \in \mathbb{N}\), then by adding we obtain \(3(a + b + c + d) = x^2 + y^2 + z^2 + t^2\), from where \[a = \frac{x^2 + y^2 + z^2 + t^2}{3} - t^2,\]\[b = \frac{x^2 + y^2 + z^2 + t^2}{3} - z^2,\]\[c = \frac{x^2 + y^2 + z^2 + t^2}{3} - y^2,\]\[d = \frac{x^2 + y^2 + z^2 + t^2}{3} - x^2.\]For these numbers to be natural and nonzero, we must choose \(x, y, z, t\) such that \(3 \mid x^2 + y^2 + z^2 + t^2\) and \(x^2 + y^2 + z^2 + t^2 > 3\max\{x^2, y^2, z^2, t^2\}\). There are many suitable choices for \(x, y, z, t\). For example, \(\{x, y, z, t\} = \{8, 9, 10, 11\}\) leads to \(\{a, b, c, d\} = \{1, 22, 41, 58\}\).