7. The sum $\sum_{i=1}^{k} a_{m+i}$ is called the continuous $k$-term sum of the sequence $a_{1}, a_{2}, \cdots, a_{n}$, where $m, k \in \mathbf{N}, k \geqslant 1, m+k \leqslant n$. The number of groups in the sequence $1,2, \cdots, 100$ where the sum of consecutive terms is a multiple of 11 is
7. 801 .

Let $S_{k}=\sum_{i=1}^{k} i=\frac{k(k+1)}{2}$.
Notice that,
$$
\begin{array}{l}
S_{k+11}=\frac{(k+11)(k+12)}{2} \\
\equiv \frac{k(k+1)}{2}=S_{k}(\bmod 11),
\end{array}
$$

and the remainders of $S_{1}, S_{2}, \cdots, S_{11}$ modulo 11 are $1,3,6$, $10,4,10,6,3,1,0,0$.

Since $100=9 \times 11+1$, the remainders of $S_{1}, S_{2}, \cdots, S_{100}$ modulo 11 include 19 ones, 18 threes, 18 sixes, 18 tens, 9 fours, and 18 zeros.
Thus, the number of qualifying groups is
$$
3 C_{18}^{2}+C_{19}^{2}+C_{18}^{2}+18=801 .
$$