The sum of the arithmetic series $ 1+2+3+ \cdots + n $ is equal to $ \frac{n(n+1)}{2} $. Let $ k $ and $ k+1 $ be the two consecutive integers removed, so that their sum is $ 2k+1 $. It follows that \[\frac{n(n + 1)}{2} - (2k+1) = 241.\] The smallest numbers that Charlize could have omitted are 1 and 2, so \[241 = \frac{n(n+1)}{2} - (2k+1) \le \frac{n(n + 1)}{2} - 3,\] which gives us the inequality $ n(n + 1) \ge 488 $. If $ n = 21 $, then $ n(n + 1) = 462 $, and if $ n = 22 $, then $ n(n + 1) = 506 $, so $ n $ must be at least 22. The largest numbers that Charlize could have omitted are $ n $ and $ n - 1 $, so \[241 = \frac{n(n+1)}{2} - (2k+1) \ge \frac{n(n + 1)}{2} - n - (n - 1) = \frac{(n - 1)(n - 2)}{2},\] which gives us the inequality $ (n - 1)(n - 2) \le 482 $. If $ n = 23 $, then $ (n - 1)(n - 2) = 462 $, and if $ n = 24 $, then $ (n - 1)(n - 2) = 506 $, so $ n $ must be at most 23. From the bounds above, we see that the only possible values of $ n $ are 22 and 23. If $ n = 22 $, then the equation \[\frac{n(n + 1)}{2} - (2k+1) = 241\] becomes $ 253 - (2k + 1) = 241 $, so $ 2k + 1 = 12 $. This is impossible, because $ 2k + 1 $ must be an odd integer. Therefore, $ n = \boxed{23} $. Note that $ n = 23 $ is possible, because Charlize can omit the numbers 17 and 18 to get the sum $ 23 \cdot 24/2 - 17 - 18 = 241 $.