Let $ t $ be the number of members of the committee, $ n_k $ be the number of votes for candidate $ k $, and let $ p_k $ be the percentage of votes for candidate $ k $ for $ k= 1,2, \dots, 27 $. We have $$n_k \ge p_k+1 = {{100n_k}\over t} +1.$$Adding these 27 inequalities yields $ t \ge 127 $. Solving for $ n_k $ gives $ n_k \ge \displaystyle{t \over{t-100}} $, and, since $ n_k $ is an integer, we obtain $$n_k \ge \biggl\lceil{t \over{t-100}}\biggr\rceil,$$where the notation $ \lceil x\rceil $ denotes the least integer that is greater than or equal to $ x $. The last inequality is satisfied for all $ k= 1,2, \dots, 27 $ if and only if it is satisfied by the smallest $ n_k $, say $ n_1 $. Since $ t \ge 27n_1 $, we obtain $$t \ge 27 \biggl\lceil{t \over {t-100}} \bigg\rceil \quad (1)$$and our problem reduces to finding the smallest possible integer $ t\ge127 $ that satisfies the inequality (1). If $ {t \over {t-100}} > 4 $, that is, $ t \le 133 $, then $ 27\left\lceil{t\over {t-100}}\right\rceil \ge27 \cdot5=135 $ so that the inequality (1) is not satisfied. Thus $ \boxed{134} $ is the least possible number of members in the committee. Note that when $ t=134 $, an election in which 1 candidate receives 30 votes and the remaining 26 candidates receive 4 votes each satisfies the conditions of the problem. $${{\bf OR}}$$ Let $ t $ be the number of members of the committee, and let $ m $ be the least number of votes that any candidate received. It is clear that $ m \ne 0 $ and $ m \ne 1 $. If $ m=2 $, then $ 2 \ge 1+100 \cdot \frac{2}{t} $, so $ t \ge 200 $. Similarly, if $ m=3 $, then $ 3 \ge 1+100 \cdot \frac{3}{t} $, and $ t \ge 150 $; and if $ m=4 $, then $ 4 \ge 1+100 \cdot \frac{4}{t} $, so $ t \ge 134 $. When $ m \ge 5 $, $ t \ge 27 \cdot 5=135 $. Thus $ t \ge 134 $. Verify that $ t $ can be $ \boxed{134} $ by noting that the votes may be distributed so that 1 candidate receives 30 votes and the remaining 26 candidates receive 4 votes each.