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Let $x = 1000a + 100b + 10c + d$ , where $a, b, c, d = {0,1,2,3,4,5,6,7,8,9}$ AND $a \neq 0.$ The problem is interested in the condition:

$1000a + 100b + 10c + d + 2088 = 1000d + 100c + 10b + a$ ;

or $-999a - 90b + 90c + 999d = 2088;$

or $111(d-a) + 10(c-b) = 232.$

The Diophantine equation $111\alpha + 10\beta = 232$ only has one solution in positive integers: $(\alpha, \beta) = (2,1).$ Hence, we require $d - a = 2$ and $c - b = 1$ which generates the ordered pairs:

$(a,d) = (1,3); (2,4); (3,5); (4,6); (5,7); (6,8); (7,9).$ , or 7 pairs

$(b,c) = (0,1); (1,2); (2,3); (3,4); (4,5); (5,6); (6,7); (7,8); (8,9)$ , or 9 pairs.

which gives a grand total of $7 \cdot 9 = \boxed{63}$ 4-digit numbers that satisfy this condition.