Let \( S_1 = \sum_{k=1}^{n}(-1)^{k-1}k = 1 - 2 + 3 - 4 + \dots + (-1)^{n-1}n \).\nIf \( n \) is even, \( S_1 = -\frac{n}{2} \).\nIf \( n \) is odd, \( S_1 = \frac{n+1}{2} \).\n\nLet \( S_2 = \sum_{k=1}^{n}(-1)^{k-1}k^2 = 1^2 - 2^2 + 3^2 - 4^2 + \dots + (-1)^{n-1}n^2 \).\nIf \( n \) is even, \( S_2 = -\frac{n(n+1)}{2} \).\nIf \( n \) is odd, \( S_2 = \frac{n(n+1)}{2} \).\n\nThe given equation is \( S_1^2 - S_2 + 2450 = 0 \).\n\nCase 1: Let \( n \) be even.\n\[ \left(-\frac{n}{2}\right)^2 - \left(-\frac{n(n+1)}{2}\right) + 2450 = 0 \]\n\[ \frac{n^2}{4} + \frac{n^2+n}{2} + 2450 = 0 \]\nThis yields all positive terms summing to zero, which is impossible since \( n > 0 \).\n\nCase 2: Let \( n \) be odd.\n\[ \left(\frac{n+1}{2}\right)^2 - \frac{n(n+1)}{2} + 2450 = 0 \]\n\[ \frac{(n+1)^2 - 2n(n+1)}{4} + 2450 = 0 \]\n\[ \frac{(n+1)(n+1 - 2n)}{4} + 2450 = 0 \]\n\[ \frac{(n+1)(1-n)}{4} = -2450 \]\n\[ \frac{1-n^2}{4} = -2450 \]\n\[ n^2 - 1 = 9800 \]\n\[ n^2 = 9801 \]\n\nTaking the square root (since \( n \) is a positive integer):\n\[ n = 99 \]\n\nThus, the correct value for \( n \) is 99.