I am confused by this question. If $X, Y$ are exchangeable and the question instead asked for $E[X | X + Y = c]$ for some constant $c$, then $E[X | X + Y = c] = c / 2$ regardless of whether $X, Y$ are independent and their individual distributions.
But the question is asking for $E[X | X + Y]$, which is a random variable. How does knowing that $X, Y$ are independent normal help?
Hint: $X$ and $Y$ are jointly normal and this implies that $X-aY$ and $X+Y$ are jointly normal too for any real number $a$. So $X-aY$ and $X+Y$ are independent if their covariance is $0$. Choose $a$ such that the covariance is $0$ [If there is no such $a$ there would be $a$ such that $Y-aX$ and $X+Y$ are independent and the proof is similar in this case]. If it so happens that $a=-1$ then $X+Y$ is independent of itself and this forces $X$ and $Y$ to be constants since $X$ and $Y$ are also given to be independent. So $1+a \neq 0$. Now $X=\frac 1 {1+a} (X-aY)+ \frac a {1+a} (X+Y)$. Now it is easy to calculate the conditional expectation.
Alternative solution:
A well known result is the following:
If $X,X_1,X_2,...,X_n$ are jointly normal then $E(X|X_1,X_2,....,X_n)$ is a linear combination of $X_1,X_2,...,X_n$.
Using this $E(X|X+Y)=c(X+Y)$ for some constant $c$ and we get $c$ from the fact that $E(X(X+Y))=cE(X+Y)^{2}$,