We know that the sum of the squares of the first \( k \) natural numbers is \( \frac{k(k+1)(2k+1)}{6} \).\n\nThus, the arithmetic mean \( A_k \) is:\n\[ A_k = \frac{1}{k} \times \frac{k(k+1)(2k+1)}{6} = \frac{(k+1)(2k+1)}{6} \]\n\[ A_k = \frac{2k^2 + 3k + 1}{6} \]\n\nSubstitute \( A_k \) into the expression:\n\[ 6A_k - 3k = (2k^2 + 3k + 1) - 3k = 2k^2 + 1 \]\n\nNow, compute the sum:\n\[ \sum_{k=1}^{n}(6A_k - 3k) = \sum_{k=1}^{n}(2k^2 + 1) \]\n\[ = 2\sum_{k=1}^{n}k^2 + \sum_{k=1}^{n}1 = 2\left(\frac{n(n+1)(2n+1)}{6}\right) + n \]\n\nWe are given that this sum equals 31. By trial and error for small natural numbers:\nIf \( n=3 \):\n\[ 2\left(\frac{3(4)(7)}{6}\right) + 3 = 2(14) + 3 = 28 + 3 = 31 \]\n\nTherefore, \( n = 3 \).