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?

Alternative way:
I think you can solve your question using generating functions such that let the exponents of following sequence represent the integers for $x_i's$ such that $$\frac{x^2}{1-x^2}=x^2+x^4+x^6+...$$
Now , we should find the coefficient of $x^{100}$ in the expansion of $$\bigg(\frac{x^2}{1-x^2}\bigg)^4$$
It is equal to find the coefficient of $x^{92}$ in the expansion of $$\bigg(\frac{1}{1-x^2}\bigg)^4 $$ , because $$\bigg(\frac{x^2}{1-x^2}\bigg)^4=(x^2)^4 \times \bigg(\frac{1}{1-x^2}\bigg)^4$$
$$[x^{92}]\bigg(\frac{1}{1-x^2}\bigg)^4=\binom{4+46-1}{3}(x^2)^{46}=18,424$$
