The layer sum of $wxyz$ equals 2014. This means that the sum of the integer with digits $wxyz$, the integer with digits $xyz$, the integer with digits $yz$, and the integer $z$ is 2014. Note that the integer with digits $wxyz$ equals $1000w + 100x + 10y + z$, the integer with digits $xyz$ equals $100x + 10y + z$, and the integer with digits $yz$ equals $10y + z$. Therefore, we have $(1000w + 100x + 10y + z) + (100x + 10y + z) + (10y + z) + z = 2014$ or $1000w + 200x + 30y + 4z = 2014$. Each of $w, x, y, z$ is a single digit and $w \neq 0$. Now $w$ cannot be 3 or greater, or the left side would be at least 3000, which is too large. Thus, $w = 1$ or $w = 2$. If $w = 2$, then $2000 + 200x + 30y + 4z = 2014$ and so $200x + 30y + 4z = 14$ or $100x + 15y + 2z = 7$. This would mean that $x = y = 0$, which gives $2z = 7$ which has no integer solutions. Thus, $w \neq 2$. Therefore, $w = 1$. This gives $1000 + 200x + 30y + 4z = 2014$ and so $200x + 30y + 4z = 1014$ or $100x + 15y + 2z = 507$. Since $0 \leq y \leq 9$ and $0 \leq z \leq 9$, then $0 \leq 15y + 2z \leq 153$. Since $100x$ is a multiple of 100 and $0 \leq 15y + 2z \leq 153$, then $100x = 400$ or $100x = 500$ so $15y + 2z = 507 - 400 = 107$ or $15y + 2z = 507 - 500 = 7$. From above, we saw that $15y + 2z$ cannot equal 7, so $15y + 2z = 107$, which means that $100x = 400$ or $x = 4$. Thus, $15y + 2z = 107$. Since $2z$ is even, then $15y$ must be odd to make $15y + 2z$ odd. The odd multiples of 15 less than 107 are $15, 45, 75, 105$. Since $0 \leq 2z \leq 18$, then we must have $15y = 105$ or $y = 7$. This gives $2z = 2$ or $z = 1$. Therefore, the integer $wxyz$ is 1471. Finally, $w + x + y + z = 1 + 4 + 7 + 1 = 13$.