Answer: 200

Solution:

Since $n$ is divisible by $5,$ its last digit is $0$ or $5;$ since the first digit equals the last digit and cannot be $0,$ both are $5.$ The outer digits contribute $10$ to the digit sum, so the three middle digits must also sum to a multiple of $5.$

Choose the second and third digits freely, in $10 \cdot 10 = 100$ ways. Whatever their sum is, the fourth digit must land in a prescribed residue class modulo $5,$ and exactly $2$ of the digits $0$ through $9$ lie in each class. The count is $10 \cdot 10 \cdot 2 = 200.$