11. Let the sum of 10 consecutive positive integers not greater than 2006 form the set $S$, and the sum of 11 consecutive positive integers not greater than 2006 form the set $T$, then the number of elements in $S \cap T$ is $\qquad$ .
11.181.
$S$ is the set of all integers ending in 5, starting from 55 to
$$
20015=\sum_{i=1}^{10}(1996+i)=10 \times 1996+55
$$

Similarly, $T$ is the set of integers starting from 66 and increasing by 11 each time, with the largest number being
$$
22011=\sum_{i=1}^{11}(1995+i)=11 \times 1995+66 \text {. }
$$
In $T$, one out of every 10 elements has a units digit of 5, so there are $199=\left[\frac{1996}{10}\right]$ elements in $T$ with a units digit of 5, which are
$$
t_{k}=55+110 k(k=1,2, \cdots, 199) .
$$

From $55+110 k \leqslant 20015$, we solve to get $k \leqslant 181$.
Thus, $|S \cap T|=181$.