Let $x_i$ be positive, even integer such that $x_1 + x_2 + x_3 + x_4 = 100.$ How many solutions are there for this expression? 
Let $x_i$ be positive, even integer such that $$x_1 + x_2 + x_3 + x_4 = 100.$$ How many solutions are there for this expression?

I think I've understood something incorrectly about the stars and bars idea. If we drop the "even" condition this has $$103\choose 3$$ solutions. Now keeping the evenness we can express $x_i=2a_i$ for $a_i$ a positive integer. We thus get $$2a_1 + 2a_2 + 2a_3 + 2 a_4 = 100$$ which is equivalent to $$a_1 + a_2 + a_3 + a_4 = 50$$ and this has $$53 \choose 3$$ solutions. However $53 \choose 3$ is not apparently the correct answer here? Is there some implication here that doesn't go backwards or what am I missing?

You are correct that the problem reduces to finding the number of positive integers $a_1, \ldots, a_4$ summing to $50$. Write $a_i = b_i + 1$ for each $i$. Then we wish to find the number of nonnegative $b_1, \ldots, b_4$ such that $b_1 + 1 + b_2 + 1 + b_3 + 1 + b_4 + 1 = 50$. That is equivalent to $b_1 + b_2 + b_3 + b_4 = 46$. Apply the stars and bars method to get the answer, $\binom{49}{3}$.
