A number is divisible by $11$ if the sum of the digits in the tens and thousands place minus the sum of the digits in the ones, hundreds, and ten thousands place is divisible by 11.
So take a number:
$a b c d e$
$(b+d)-(a+c+e)$ is divisible by $11$ (it may be $0$).
$a+b+c+d+e=30$.
$a,b,c,d,e$ are all integers that may be $0$ through $9$. $a\not=0$.
Combining the two equations above, $2(b+d)-30$ is divisible by $11$, but is also even (since $b$ and $d$ are integers), so it may be $0$, $22$, or $-22$.
If it is $22$, then $(b+d)=26$, which is a contradiction.
If it is $0$, then $(b+d)=(a+c+e)=15$ (condition 1)
If it is $-22$, then $(b+d)=4$ and $(a+c+e)=26$ (condition 2)
Take all such $a,b,c,d,e$ that satisfy the above statements.
If condition 1 holds, then there are $4$ different ways to choose $b$ and $d$ and $4+5+6+7+8+9+9+8+7+6=69$ ways to choose $a,c,e$ that satisfy $a+c+e=15$. Thus, for condition 1, there are $276$ distinct numbers that satisfy condition 1.
If condition 2 holds, then there are $4$ different ways to choose $b$ and $d$ and $3$ ways to choose $a,c,e$. So there are $12$ distinct numbers that satisfy condition 2.
Thus, there are $288$ distinct 5-digit numbers that satisfy the desired condition.
Best Answer
If we look at the number $a_6a_5a_4a_3a_2a_1a_0$: $$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$ So if the number will be divisible by 11 we require $$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$
We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to: $$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$ Which really means that $11m \in \{-22,-11,0,11,22,33\}$
The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:
$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2 \\a_5+a_3+a_1 =\frac{ 59-11m}2$$
Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:
$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$ Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly $$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with $$a_6 +a_4+a_2+a_0 = 35 \\a_5+a_3+a_1 =24$$
It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.
Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.