$X = \Bbb{CP}^4 \#\Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.
Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $\Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;\Bbb Z)$ as $x_1, x_2$.
1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.
2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.
3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $\Bbb{CP}^2 \vee \Bbb{CP}^2$. Because $X_4 \hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X \to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f \vee f$, where $f$ is the restriction of the classifying map of $\Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(\Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.
Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2\ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).
Then because $\chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+\ell n) + 16(m+n) - (2k+1)^4 - (2\ell+1)^4 + 10(2k+1)^2 + 10(2\ell+1)^2 - 50.$$
Simplifying we get $$80 = 32(km + \ell n - k^3 - \ell^3 + k^2 + \ell^2) + 16(k+\ell - k^4 - \ell^4 +m +n)$$
Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 \equiv 16(k+\ell - k^4- \ell^4) \mod 32$$ But $k+\ell$ is odd iff $k^4 + \ell^4$ is odd, so the right side is $0 \mod 32$. This is a contradiction, as desired.
That this was so much work suggests that this is, uh, the wrong approach to prove that $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.
There are tons and tons of examples: $\mathbb{CP}^n$ for $n \ge 2$, $S^n \times S^m$ for $n,m \ge 2$, connected sums of these when they have the same dimensions, products of these... The list goes on and on.
It might help to review why a 2- or 3-dimensional closed manifold which is simply connected is a sphere to see why it fails in higher dimensions.
In dimension 2 it follows from the classification of closed surfaces, and there is no classification of manifolds in higher dimension that could tell us anything like that. This isn't really insightful.
Now in dimension 3. Hurewicz's theorem implies that $\require{cancel}H_1(M) \cong \cancel{\pi_1(M)}_{\mathrm{ab}} = 0$ because $M$ is simply connected. A simply connected manifold is always orientable, and so you can apply Poincaré duality $H^2(M) \cong H_1(M) = 0$, and the universal coefficients theorem implies $H_2(M) = 0$ (I'm glossing a bit over the details here, you also need the fact that every homology group of a closed manifold is finitely generated). So you're only left with $H_0(M) \cong H_3(M) \cong \mathbb{Z}$, in other words your manifold is a homology sphere.
Again Hurewicz's theorem implies $\pi_3(M) \cong \mathbb{Z}$, so there's a representative of a generator $f : S^3 \to M$ that induces an isomorphism on $\pi_3$. This $f$ then induces an isomorphism on $H_3$, and since there is no other nontrivial homology, $f$ is a homology equivalence. Since both spaces are simply connected, $f$ has to be a weak homotopy equivalence, and by Whitehead's theorem a homotopy equivalence. (Showing that the manifold is actually homeomorphic to a sphere would have earned you a million dollars back in the day.)
So what fails in higher dimension? Poincaré duality is not enough anymore to ensure that all homology beyond the top homology group vanishes. So you can have nontrivial $H_2$, and then all bets are off. But in dimension 3, there's just not enough room in (co)homology, so the manifold has to be a homology sphere; but a simply connected homology sphere is always homotopy equivalent to a sphere.
Best Answer
Some points.
By Poincare duality the Euler characteristic of a closed odd-dimensional manifold vanishes so we restrict our attention to even-dimensional manifolds. By the classification of surfaces $S^2$ is the only example in dimension $2$.
Euler characteristic is multiplicative with respect to products (e.g. by the Kunneth theorem). So we can find examples of manifolds with positive Euler characteristic by taking products of an even number of manifolds of negative Euler characteristic together with any number of manifolds of positive Euler characteristic. The Euler characteristic is also multiplicative with respect to (nice) fiber bundles so we can consider nontrivial fiber bundles with suitable bases and fibers also.
The Euler characteristic of a connected sum of closed $n$-manifolds satisfies the "inclusion-exclusion" formula $\chi(M\#N) = \chi(M) + \chi(N) - \chi(S^n)$. This means when $n$ is even, connected sum with $N$ increases the Euler characteristic iff $\chi(N) \ge 3$. Also the connected sum of simply connected manifolds is simply connected.
Any manifold whose cohomology is concentrated in even degrees (say, over $\mathbb{Q}$) has positive Euler characteristic, and there's a large source of examples of such manifolds coming from algebraic geometry: every generalized flag variety $G/P$ over $\mathbb{C}$ has this property. $\mathbb{CP}^n$ is a special case of this construction but we also have Grassmannians and complete flag varieties, for example, which are also simply connected (probably generalized flag varieties over $\mathbb{C}$ are always simply connected but I don't know how to prove it).
By Poincare duality, any closed simply connected $4$-manifold has cohomology concentrated in even degrees and hence has Euler characteristic at least $2$. This blog post on hypersurfaces in $\mathbb{CP}^3$ discusses their topology; in particular they are completely classified up to homotopy (Milnor, Whitehead) and even up to homeomorphism (Freedman). Their Euler characteristics can be arbitrarily large: the linked post shows that the Euler characteristic of a smooth degree $d$ hypersurface in $\mathbb{CP}^3$ is $d^3 - 4d^2 + 6d$. When $d = 1$ we get $\mathbb{CP}^2$ which has $\chi = 1 - 4 + 6 = 3$, when $d = 2$ we get $\mathbb{CP}^1 \times \mathbb{CP}^1$ which has $\chi = 8 - 16 + 12 = 4$, and when $d = 4$ we get a K3 surface which has $\chi = 64 - 64 + 24 = 24$.
For the classification of closed simply connected $6$-manifolds see here. Wall showed that any such manifold splits as a connected sum of a number of copies of $S^3 \times S^3$ ($\chi(S^3 \times S^3) = 0$ so this connected sum lowers the Euler characteristic by $2$ and removing it increases the Euler characteristic by $2$) and a manifold $M$ with $b_3 = 0$, hence $M$ has rational cohomology concentrated in even degrees and so has positive Euler characteristic. Incidentally, this makes a connected sum of two copies of $S^3 \times S^3$ the simplest example of a closed simply connected manifold with negative Euler characteristic, so now we can actually carry out the construction I suggested in point 2: the product of two such sums is a closed simply connected $12$-manifold with positive Euler characteristic whose rational cohomology is not concentrated in even degree.
The Hopf conjecture claims in part that a closed even-dimensional manifold admitting a metric with positive sectional curvature has positive Euler characteristic. In dimension $2$ this of course follows from the Gauss-Bonnet theorem, and as Wikipedia discusses this also holds in dimension $4$.
If I were looking for more examples I might look through lists of symmetric spaces.