R, 63 59 bytes

s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])

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-4 bytes thanks to Giuseppe

(The explanation contains a spoiler as to how to solve the problem without computing $k$.)

Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. There exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:

• s[2]<2 and s[2]!=s[1]


• s[3]<3 and s[3]!=s[2]


• s[4]<5 and s[4]!=s[3]

The code can probably be golfed further.

Haskell, 69 bytes

Based on the Chinese remainder theorem

m=[2,3,5,7]

f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]

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C# (Visual C# Interactive Compiler), 125 bytes

n=>Enumerable.Range(n+1,209).Any(c=>new{c%2,c%3,c%5,c%7}.OrderBy(x=>x).SequenceEqual(new{n%2,n%3,n%5,n%7}.OrderBy(x=>x)))

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Wolfram Language (Mathematica), 67 bytes

!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&

R@n_:=n~Mod~{2,3,5,7}

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Perl 6, 64 61 59 bytes

{my&f={bag($_ X%2,3,5,7)};any(($_+1+ ^209).map(&f))===f $_}

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Fixed my bug by changing set to bag :).

Explanation:

{my&f={bag($_ X%2,3,5,7)};any(($_+1+ ^209).map(&f))===f $_}

 my&f={bag($_ X%2,3,5,7)};                                  # helper function: f

      {    $_           };                                  # takes the implicit argument $_

              X%2,3,5,7                                     # cross metaoperator with divide: does cross product division

       bag(            )                                    # stores the results in a bag (weighed set, knows how many identical items it contains)

                              ($_+1+ ^209)                  # Generates a range of the argument +1 up to 209+argument+1

                                          .map(&f)          # Apply the helper function for each value in that range

                          any(                    )         # Is any of these bags

                                                   ===      # Equal to

                                                      f $_  # The helper function result for the given argument?

Alternative version:

{my @a=($_+ ^209).map:{bag($_ X%2,3,5,7)};@a.shift===any @a}

Ruby, 54 bytes

->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}

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Uses Robin Ryder's clever solution.

R, 72 bytes

n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F

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Jelly, 15 bytes

8ÆR©PḶ+%Ṣ¥€®ċḢ$

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I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.

PHP, 81 78 72 bytes

while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);

A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.

$ echo 3|php -nF euc.php

T

$ echo 5|php -nF euc.php

T

$ echo 2019|php -nF euc.php

T

$ echo 0|php -nF euc.php



$ echo 2|php -nF euc.php



$ echo 1999|php -nF euc.php

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Or 73 bytes with 1 or 0 response

while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;

$ echo 2019|php -nF euc.php

1

$ echo 1999|php -nF euc.php

0

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Original answer, 133 127 bytes

function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}

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Python 2, 41 bytes

lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4

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Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing the the three conditions turned out shorter than writing a general one:

46 bytes

lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])

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Python 3, 69 bytes

lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210

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Hardcoded