Definitions, Terms, and Notations: For any number $ X $ with base-$ 10 $ expansion $ d_{1} 10^{n} + d_{2} 10^{n-1} + ... + d_{n+1} $, where $ d_{1}, d_{2}, ... d_{n+1} $ are its digits, call the digit $ d_{i} $ lefter than digit $ d_{j} $ in the number $ X $ if $ i < j $ for integers $ 1 \le i, j \le n+1 $. For the sake of simplicity, we introduce the notation $ d_{1} d_{2} d_{3} ... d_{n} $ to represent the number any number $ X $ with those digits. Let $ a_{1}, a_{2}, ..., a_{7} $ be the seven deleted digits of one of the numbers (call this number $ N_{1} $). $ a_{i} $ is lefter than $ a_{j} $ in the number $ N_{2} $ if $ i < j $. Let $ b_{1}, b_{2}, ..., b_{7} $ be the seven deleted digits of the other number (call this number $ N_{2} $). $ b_{i} $ is lefter than $ b_{j} $ in the number $ N_{2} $ if $ i < j $. Let $ n_{1} n_{2} n_{3} ... n_{1999} n_{2000} $ be the number formed after deleting the seven digits of $ N_{1} $ or $ N_{2} $. Note that the digits $ n_{1}, n_{2}, n_{3}, ..., n_{2000} $ are also digits of $ N_{1} $ and $ N_{2} $. Thus, $ b_{i} $ is lefter than some of those digits in the number $ N_{2} $ for integer $ 1 \le i \le 7 $. Let $ b_{i} $ be lefter than $ n_{j_{i}} $, but $ n_{j_{i} + 1} $ is lefter than $ b_{i} $ in the number $ N_{2} $ for all $ i = 1, ..., 7 $, unless $ b_{1} $ is the leftmost digit of $ N_{2} $ (also, $ 1 \le j_{i} \le 2000 $ and are integral). If $ b_{1} $ is the leftmost digit, than it is simply lefter than all the digits of the number $ N_{2} $. Solution: In the number $ N_{1} $, $ n_{1}, n_{2}, ... n_{2000} $ are also digits. Thus, construct a new number $ N_{3} $ by inserting $ b_{i} $ in between the digits $ n_{j_{i}} $ and $ n_{j_{i} + 1} $. Notice that by deleting the digits $ a_{1}, a_{2}, ..., a_{7} $, $ N_{3} $ becomes the number $ N_{2} $. Thus, we can add the digits $ a_{1}, a_{2}, ..., a_{7} $ to $ N_{2} $ to form the number $ N_{3} $. Hence, it is possible to add $ 7 $ digits to each of those numbers to form the same $ 2014 $-digit number.

Meh, too hard to actually write up a solution for this. Some of the notations, definitions, and terms weren't really used and the solution might've used some other terms that should've been written. The idea is easy to visualize, but any problems involving digits are just too difficult to write solutions for. In short, the solution says: just insert the same seven digits into each number and in the right spots.