易知，要使\(f(n)\)取得最小值，正整数n必然在区间\([1,20]\)上，
则\(f(n) = (n - 1) + (n - 2) + \cdots + 3 + 2 + 1 + 0 + 1 + 2 + 3 + \cdots + (20 - n)\)\( = \frac{{(n - 1)[1 + (n - 1)]}}{2} + \frac{{(20 - n)[1 + (20 - n)]}}{2}\)\( = {n^2} - 21n + 210\)\( = {\left( {n - \frac{{21}}{2}} \right)^2} + \frac{{399}}{4}\)
∵\(n \in {{\mathbf{N}}_ + }\)，∴\(n = 10\)或\(n = 11\)时\(f(n)\)有最小值\(100\)．
故答案为：100
