Solution:
For a number $d_{5}d_{4}d_{3}d_{2}d_{1}d_{0}$ (allowing leading 0s) to be upwards and a multiple of 11, we must have
$$d_{5} \leq d_{4} \leq d_{3} \leq d_{2} \leq d_{1} \leq d_{0},$$
$$d_{0} - d_{1} + d_{2} - d_{3} + d_{4} - d_{5} \equiv 0 \pmod{11}.$$
Note that $d_{0} - d_{1}$, $d_{2} - d_{3}$, and $d_{4} - d_{5}$ are all nonnegative. Thus,
$$0 \leq (d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5})$$
$$\leq (d_{0} - d_{1}) + (d_{1} - d_{2}) + (d_{2} - d_{3}) + (d_{3} - d_{4}) + (d_{4} - d_{5})$$
$$= d_{0} - d_{5}$$
$$\leq 9.$$
Therefore,
$$(d_{0} - d_{1}) + (d_{2} - d_{3}) + (d_{4} - d_{5}) = 0,$$
which can only occur when $d_{0} = d_{1}$, $d_{2} = d_{3}$, and $d_{4} = d_{5}$, i.e. the number is of the form $aabbcc$. We can easily verify that all numbers of the form $aabbcc$ for digits $a \leq b \leq c$ satisfy our conditions, so we simply have to count them.
There are $\binom{12}{3} = 220$ such triples of digits $(a,b,c)$. However, one of these triples is $(0,0,0)$, which corresponds to the number 0. Thus our answer is $220 - 1 = 219$.