Let $x_1, x_2, \ldots, x_n$ be non-negative real numbers such that $x_1 + x_2 + \cdots + x_n = 1$. We aim to find the largest possible value of $\sum_{j=1}^n (x_j^4 - x_j^5)$.

To solve this, we use the method of smoothing. We start by considering small cases and then generalize.

### Key Claim:
If $x + y < \frac{7}{10}$, then:
$$(x + y)^4 - (x + y)^5 > x^4 - x^5 + y^4 - y^5.$$

### Proof of the Claim:
Consider the inequality:
$$(x + y)^4 - (x + y)^5 > x^4 - x^5 + y^4 - y^5.$$
Expanding and simplifying both sides, we get:
$$4x^2 + 4y^2 + 6xy > 5x^3 + 5y^3 + 10x^2y + 10xy^2.$$
Rewriting the left-hand side (LHS) and right-hand side (RHS), we have:
$$\text{LHS} = \frac{7}{2}(x^2 + y^2) + \frac{1}{2}(x^2 + y^2) + 6xy \geq \frac{7}{2}(x + y)^2,$$
$$\text{RHS} \leq 5(x^3 + y^3 + 3x^2y + 3xy^2) = 5(x + y)^3.$$
Thus, if $x + y < \frac{7}{10}$, the inequality holds.

### General Case:
Let $k$ be the number of non-zero $x_j$ among $x_1, \ldots, x_n$. Without loss of generality, assume:
$$x_1 \geq x_2 \geq \cdots \geq x_k > 0, \quad x_{k+1} = x_{k+2} = \cdots = x_n = 0.$$
If $k \geq 3$, denote:
$$x_i' = x_i \quad (i = 1, 2, \ldots, k-2), \quad x_{k-1}' = x_{k-1} + x_k, \quad x_k' = x_{k+1}' = \cdots = x_n' = 0.$$
Since $x_{k-1} + x_k \leq \frac{2}{n} \leq \frac{2}{3} < \frac{7}{10}$, by the claim, we have:
$$\sum_{j=1}^n (x_j'^4 - x_j'^5) > \sum_{j=1}^n (x_j^4 - x_j^5).$$
This smoothing process can be repeated until at most two $x_j$ are non-zero.

### Final Step:
Let $x_1 = a$ and $x_2 = b$ with $a + b = 1$. Then:
$$S = a^4 - a^5 + b^4 - b^5 = ab(a^3 + b^3) = ab(a + b)(a^2 + b^2 - ab) = ab(1 - 3ab).$$
Maximizing $S$, we find:
$$S \leq \frac{1}{12}.$$
Equality holds when $a = \frac{3 + \sqrt{3}}{6}$ and $b = \frac{3 - \sqrt{3}}{6}$.

The answer is: \boxed{\frac{1}{12}}.