2 fixed a small inaccuracy

Dear Georges, maybe the following argument works. (It's quite possible a sign went wrong somewhere, though.)

Let $\pi: Y \rightarrow X$ be the blowup. By assumption Y is projective, so it carries an ample line bundle A say. Let E denote the exceptional divisor of the blowup, and consider line bundles of the form $A+nE$ (for positive integers n). If C is any curve in Y which is not contained in E, then $(A+nE).C = A.C + nE.C$ is positive, for any n. On the other hand, let L be a line in E: then $E.L = -1$. (Note that all other curves in E are numerical multiples of this one.) So if we set n=A.L (positive, by ampleness of A) then we have $(A+nE).L = 0$. So the line bundle A+nE is nef, and has degree 0 exactly on those curves which lie in E.

I claim that $A+nE$ is in fact basepoint-free. To see this, it suffices (by the Basepoint-free Theorem, see e.g. Koll\'ar--Mori Chapter 3) to show that the line bundle $m(A+nE)-K_Y$ is nef and big, for some positive integer m. Now A is ample and E is effective, so $A+nE$ is big for all positive n (ample+effective=big --- this is also in Koll\'ar--Mori). Moreover, bigness is an open condition, so for m sufficiently large, $m(A+nE)-K_Y$ is still big. So it remains to prove nefness.

Recall that we were free to choose A to be any ample line bundle, so choose it to satisfy the condition that $A-K_Y$ is itself ample (again, using the fact that ampleness is open). Then $m(A+nE)-K_Y = (mA-K_Y) +mnE$, so in particular it has positive degree on any curve C which is not contained in E. On the other hand, if L is a line in E, then $(m(A+nE)-K_Y).L = -K_Y.L$ (by the calculations in the first pgph.)paragraph.). Moreover, $-K_Y=\pi^\ast(-K_X)-E$, -K_Y=\pi^\ast(-K_X)-(dim X-1)E$, so$-K_Y.L=-E.L=1$. -K_Y.L=-(dim X-1)E.L>0$. So $m(A+nE)-K_Y$ has nonnegative degree on all curves, i.e. it is nef.

Putting all this together, we have that $m(A+nE)$ is basepoint-free for suitable positive integers m and n. So it defines a contraction morphism $p: Y \rightarrow Z$ to another projective variety Z. But the morphism p contracts exactly those curves on which A+nE has degree 0, which by construction are exactly the curves contained in E. Therefore Z is exactly the blow-down of E to a point, hence isomorphic to X. Since Z is projective, so is X.

1

Dear Georges, maybe the following argument works. (It's quite possible a sign went wrong somewhere, though.)

Let $\pi: Y \rightarrow X$ be the blowup. By assumption Y is projective, so it carries an ample line bundle A say. Let E denote the exceptional divisor of the blowup, and consider line bundles of the form $A+nE$ (for positive integers n). If C is any curve in Y which is not contained in E, then $(A+nE).C = A.C + nE.C$ is positive, for any n. On the other hand, let L be a line in E: then $E.L = -1$. (Note that all other curves in E are numerical multiples of this one.) So if we set n=A.L (positive, by ampleness of A) then we have $(A+nE).L = 0$. So the line bundle A+nE is nef, and has degree 0 exactly on those curves which lie in E.

I claim that $A+nE$ is in fact basepoint-free. To see this, it suffices (by the Basepoint-free Theorem, see e.g. Koll\'ar--Mori Chapter 3) to show that the line bundle $m(A+nE)-K_Y$ is nef and big, for some positive integer m. Now A is ample and E is effective, so $A+nE$ is big for all positive n (ample+effective=big --- this is also in Koll\'ar--Mori). Moreover, bigness is an open condition, so for m sufficiently large, $m(A+nE)-K_Y$ is still big. So it remains to prove nefness.

Recall that we were free to choose A to be any ample line bundle, so choose it to satisfy the condition that $A-K_Y$ is itself ample (again, using the fact that ampleness is open). Then $m(A+nE)-K_Y = (mA-K_Y) +mnE$, so in particular it has positive degree on any curve C which is not contained in E. On the other hand, if L is a line in E, then $(m(A+nE)-K_Y).L = -K_Y.L$ (by the calculations in the first pgph.). Moreover, $-K_Y=\pi^\ast(-K_X)-E$, so $-K_Y.L=-E.L=1$. So $m(A+nE)-K_Y$ has nonnegative degree on all curves, i.e. it is nef.

Putting all this together, we have that $m(A+nE)$ is basepoint-free for suitable positive integers m and n. So it defines a contraction morphism $p: Y \rightarrow Z$ to another projective variety Z. But the morphism p contracts exactly those curves on which A+nE has degree 0, which by construction are exactly the curves contained in E. Therefore Z is exactly the blow-down of E to a point, hence isomorphic to X. Since Z is projective, so is X.