Since $91=7 \times 13$, we see that the possible values of \operatorname{gcd}(n, 91) are 1, 7, 13, 91. For $1 \leq n \leq 91$, there is only one value of $n$ such that \operatorname{gcd}(n, 91)=91. Then, we see that there are 12 values of $n$ for which \operatorname{gcd}(n, 91)=7 (namely, multiples of 7 other than 91 ), 6 values of $n$ for which \operatorname{gcd}(n, 91)=13 (the multiples of 13 other than 91 ), and $91-1-6-12=72$ values of $n$ for which \operatorname{gcd}(n, 91)=1. Hence, our answer is $1 \times 91+12 \times 7+6 \times 13+72 \times 1=325$.