7. The sum $\sum_{i=1}^{k} a_{m+i}$ is called the sum of $k$ consecutive terms 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 of consecutive terms in the sequence $1,2, \cdots, 100$ whose sum is a multiple of 11 is $\qquad$.
7.801.

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

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 $100=9 \times 11+1$, thus, among the remainders of $S_{1}, S_{2}, \cdots, S_{100}$ modulo 11, there are 19 ones, 18 threes, 18 sixes, 18 tens, 9 fours, and 18 zeros.
Therefore, the number of qualifying groups is
$$
3 \mathrm{C}_{18}^{2}+\mathrm{C}_{19}^{2}+\mathrm{C}_{18}^{2}+18=801 .
$$